Maximum length perimeter of a box whose diagonal is 10 unit long.












2












$begingroup$


Suppose the diagonal of a three-dimentional box has length $10$, what is its maximum perimeter length?



Here is my solution:



Let the three edges of the box adjacent to a vertex be labelled $a,b,c$. Then by Cauchy-Schwarz inequality, we have the size of the perimeters of the box is $$4 cdot langle(1,1,1),(a,b,c)rangle le 4 sqrt{a^2+b^2+c^2}sqrt{1^2+1^2+1^2}=4 cdot 10 cdot sqrt3.$$ Hence we know that the size of the perimeter of the box is bounded by $40sqrt3$. Then I still need to find an actual box with perimeter that value and with the size of its diagonal equals to $10$.






calculus geometry maxima-minima cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Suppose the diagonal of a three-dimentional box has length $10$, what is its maximum perimeter length?



    Here is my solution:



    Let the three edges of the box adjacent to a vertex be labelled $a,b,c$. Then by Cauchy-Schwarz inequality, we have the size of the perimeters of the box is $$4 cdot langle(1,1,1),(a,b,c)rangle le 4 sqrt{a^2+b^2+c^2}sqrt{1^2+1^2+1^2}=4 cdot 10 cdot sqrt3.$$ Hence we know that the size of the perimeter of the box is bounded by $40sqrt3$. Then I still need to find an actual box with perimeter that value and with the size of its diagonal equals to $10$.






    calculus geometry maxima-minima cauchy-schwarz-inequality







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Suppose the diagonal of a three-dimentional box has length $10$, what is its maximum perimeter length?



      Here is my solution:



      Let the three edges of the box adjacent to a vertex be labelled $a,b,c$. Then by Cauchy-Schwarz inequality, we have the size of the perimeters of the box is $$4 cdot langle(1,1,1),(a,b,c)rangle le 4 sqrt{a^2+b^2+c^2}sqrt{1^2+1^2+1^2}=4 cdot 10 cdot sqrt3.$$ Hence we know that the size of the perimeter of the box is bounded by $40sqrt3$. Then I still need to find an actual box with perimeter that value and with the size of its diagonal equals to $10$.






      calculus geometry maxima-minima cauchy-schwarz-inequality







      share|cite|improve this question











      $endgroup$




      Suppose the diagonal of a three-dimentional box has length $10$, what is its maximum perimeter length?



      Here is my solution:



      Let the three edges of the box adjacent to a vertex be labelled $a,b,c$. Then by Cauchy-Schwarz inequality, we have the size of the perimeters of the box is $$4 cdot langle(1,1,1),(a,b,c)rangle le 4 sqrt{a^2+b^2+c^2}sqrt{1^2+1^2+1^2}=4 cdot 10 cdot sqrt3.$$ Hence we know that the size of the perimeter of the box is bounded by $40sqrt3$. Then I still need to find an actual box with perimeter that value and with the size of its diagonal equals to $10$.






      calculus geometry maxima-minima cauchy-schwarz-inequality




      calculus geometry maxima-minima cauchy-schwarz-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 10:46









      GNUSupporter 8964民主女神 地下教會

      12.8k72445




      12.8k72445










      asked Nov 13 '18 at 19:41









      nafhgoodnafhgood

      1,805422




      1,805422






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=sqrt{frac{100}{3}}$.



          To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+b+c$ and $g(a,b,c)=a^2+b^2+c^2-100$. Then we have to find $lambda,a,b,c$ such that the Lagrange function



          $$left(frac{partial f}{partial a},frac{partial f}{partial b},frac{partial f}{partial c}right)-lambdaleft(frac{partial g}{partial a},frac{partial g}{partial b},frac{partial g}{partial c}right)$$
          is zero. This gives
          $$(1,1,1)-lambda(2a,2b,2c)=0$$
          So $$a=b=c=frac{1}{2lambda}$$
          The constraint $g(a,b,c)=0$ gives us $lambda=pmsqrt{frac{3}{400}}$, and we get the solutions
          $$(a,b,c)=left(sqrt{frac{100}{3}},sqrt{frac{100}{3}},sqrt{frac{100}{3}}right)$$
          and
          $$(a,b,c)=left(-sqrt{frac{100}{3}},-sqrt{frac{100}{3}},-sqrt{frac{100}{3}}right)$$



          corresponding to the maximum and minimum values of $a+b+c$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Your optimal solution gives $a^2 + b^2 + c^2 = 10$ instead of $100$.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:15










          • $begingroup$
            @GNUSupporter8964民主女神地下教會: Thank you for that. I have fixed it now.
            $endgroup$
            – TonyK
            Dec 11 '18 at 10:20










          • $begingroup$
            Given that OP has already found an upper bound by Cauchy-Schwarz, it suffices to find the values of $a$, $b$, $c$ whose sum attains that upper bound to finish the question. Anyways, it's pedagical to use Lagrange multipliers.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:33



















          1












          $begingroup$

          With no existing correct answer, I feel I have to add one. Actually, that's a classic inequalities problem that doesn't even need calculus, and has a one-line solution.



          By $s$-power inequality, $4(a+b+c) le 4 cdot 3 cdot sqrt{dfrac{a^2+b^2+c^2}{3}} = 40 sqrt3$. Equality holds iff $a = b = c = sqrt{dfrac{10^2}{3}} = dfrac{10 sqrt3}{3}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, the am-gm inequality says that :$$frac{x_1+ ldots + x_n}{n} geq sqrt[n]{x_1 cdots x_n}$$. How does this gives the above result?
            $endgroup$
            – nafhgood
            Dec 11 '18 at 10:39










          • $begingroup$
            @mathnoob Sorry for the name confusion. It should be called the $s$-power inequality (called "generalized mean inequality" in Wiki. AM-GM is just a special case of that inequality.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:43











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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=sqrt{frac{100}{3}}$.



          To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+b+c$ and $g(a,b,c)=a^2+b^2+c^2-100$. Then we have to find $lambda,a,b,c$ such that the Lagrange function



          $$left(frac{partial f}{partial a},frac{partial f}{partial b},frac{partial f}{partial c}right)-lambdaleft(frac{partial g}{partial a},frac{partial g}{partial b},frac{partial g}{partial c}right)$$
          is zero. This gives
          $$(1,1,1)-lambda(2a,2b,2c)=0$$
          So $$a=b=c=frac{1}{2lambda}$$
          The constraint $g(a,b,c)=0$ gives us $lambda=pmsqrt{frac{3}{400}}$, and we get the solutions
          $$(a,b,c)=left(sqrt{frac{100}{3}},sqrt{frac{100}{3}},sqrt{frac{100}{3}}right)$$
          and
          $$(a,b,c)=left(-sqrt{frac{100}{3}},-sqrt{frac{100}{3}},-sqrt{frac{100}{3}}right)$$



          corresponding to the maximum and minimum values of $a+b+c$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Your optimal solution gives $a^2 + b^2 + c^2 = 10$ instead of $100$.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:15










          • $begingroup$
            @GNUSupporter8964民主女神地下教會: Thank you for that. I have fixed it now.
            $endgroup$
            – TonyK
            Dec 11 '18 at 10:20










          • $begingroup$
            Given that OP has already found an upper bound by Cauchy-Schwarz, it suffices to find the values of $a$, $b$, $c$ whose sum attains that upper bound to finish the question. Anyways, it's pedagical to use Lagrange multipliers.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:33
















          3












          $begingroup$

          You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=sqrt{frac{100}{3}}$.



          To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+b+c$ and $g(a,b,c)=a^2+b^2+c^2-100$. Then we have to find $lambda,a,b,c$ such that the Lagrange function



          $$left(frac{partial f}{partial a},frac{partial f}{partial b},frac{partial f}{partial c}right)-lambdaleft(frac{partial g}{partial a},frac{partial g}{partial b},frac{partial g}{partial c}right)$$
          is zero. This gives
          $$(1,1,1)-lambda(2a,2b,2c)=0$$
          So $$a=b=c=frac{1}{2lambda}$$
          The constraint $g(a,b,c)=0$ gives us $lambda=pmsqrt{frac{3}{400}}$, and we get the solutions
          $$(a,b,c)=left(sqrt{frac{100}{3}},sqrt{frac{100}{3}},sqrt{frac{100}{3}}right)$$
          and
          $$(a,b,c)=left(-sqrt{frac{100}{3}},-sqrt{frac{100}{3}},-sqrt{frac{100}{3}}right)$$



          corresponding to the maximum and minimum values of $a+b+c$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Your optimal solution gives $a^2 + b^2 + c^2 = 10$ instead of $100$.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:15










          • $begingroup$
            @GNUSupporter8964民主女神地下教會: Thank you for that. I have fixed it now.
            $endgroup$
            – TonyK
            Dec 11 '18 at 10:20










          • $begingroup$
            Given that OP has already found an upper bound by Cauchy-Schwarz, it suffices to find the values of $a$, $b$, $c$ whose sum attains that upper bound to finish the question. Anyways, it's pedagical to use Lagrange multipliers.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:33














          3












          3








          3





          $begingroup$

          You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=sqrt{frac{100}{3}}$.



          To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+b+c$ and $g(a,b,c)=a^2+b^2+c^2-100$. Then we have to find $lambda,a,b,c$ such that the Lagrange function



          $$left(frac{partial f}{partial a},frac{partial f}{partial b},frac{partial f}{partial c}right)-lambdaleft(frac{partial g}{partial a},frac{partial g}{partial b},frac{partial g}{partial c}right)$$
          is zero. This gives
          $$(1,1,1)-lambda(2a,2b,2c)=0$$
          So $$a=b=c=frac{1}{2lambda}$$
          The constraint $g(a,b,c)=0$ gives us $lambda=pmsqrt{frac{3}{400}}$, and we get the solutions
          $$(a,b,c)=left(sqrt{frac{100}{3}},sqrt{frac{100}{3}},sqrt{frac{100}{3}}right)$$
          and
          $$(a,b,c)=left(-sqrt{frac{100}{3}},-sqrt{frac{100}{3}},-sqrt{frac{100}{3}}right)$$



          corresponding to the maximum and minimum values of $a+b+c$.






          share|cite|improve this answer











          $endgroup$



          You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=sqrt{frac{100}{3}}$.



          To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+b+c$ and $g(a,b,c)=a^2+b^2+c^2-100$. Then we have to find $lambda,a,b,c$ such that the Lagrange function



          $$left(frac{partial f}{partial a},frac{partial f}{partial b},frac{partial f}{partial c}right)-lambdaleft(frac{partial g}{partial a},frac{partial g}{partial b},frac{partial g}{partial c}right)$$
          is zero. This gives
          $$(1,1,1)-lambda(2a,2b,2c)=0$$
          So $$a=b=c=frac{1}{2lambda}$$
          The constraint $g(a,b,c)=0$ gives us $lambda=pmsqrt{frac{3}{400}}$, and we get the solutions
          $$(a,b,c)=left(sqrt{frac{100}{3}},sqrt{frac{100}{3}},sqrt{frac{100}{3}}right)$$
          and
          $$(a,b,c)=left(-sqrt{frac{100}{3}},-sqrt{frac{100}{3}},-sqrt{frac{100}{3}}right)$$



          corresponding to the maximum and minimum values of $a+b+c$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 10:19

























          answered Nov 14 '18 at 13:24









          TonyKTonyK

          42.5k355134




          42.5k355134








          • 1




            $begingroup$
            Your optimal solution gives $a^2 + b^2 + c^2 = 10$ instead of $100$.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:15










          • $begingroup$
            @GNUSupporter8964民主女神地下教會: Thank you for that. I have fixed it now.
            $endgroup$
            – TonyK
            Dec 11 '18 at 10:20










          • $begingroup$
            Given that OP has already found an upper bound by Cauchy-Schwarz, it suffices to find the values of $a$, $b$, $c$ whose sum attains that upper bound to finish the question. Anyways, it's pedagical to use Lagrange multipliers.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:33














          • 1




            $begingroup$
            Your optimal solution gives $a^2 + b^2 + c^2 = 10$ instead of $100$.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:15










          • $begingroup$
            @GNUSupporter8964民主女神地下教會: Thank you for that. I have fixed it now.
            $endgroup$
            – TonyK
            Dec 11 '18 at 10:20










          • $begingroup$
            Given that OP has already found an upper bound by Cauchy-Schwarz, it suffices to find the values of $a$, $b$, $c$ whose sum attains that upper bound to finish the question. Anyways, it's pedagical to use Lagrange multipliers.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:33








          1




          1




          $begingroup$
          Your optimal solution gives $a^2 + b^2 + c^2 = 10$ instead of $100$.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 11 '18 at 10:15




          $begingroup$
          Your optimal solution gives $a^2 + b^2 + c^2 = 10$ instead of $100$.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 11 '18 at 10:15












          $begingroup$
          @GNUSupporter8964民主女神地下教會: Thank you for that. I have fixed it now.
          $endgroup$
          – TonyK
          Dec 11 '18 at 10:20




          $begingroup$
          @GNUSupporter8964民主女神地下教會: Thank you for that. I have fixed it now.
          $endgroup$
          – TonyK
          Dec 11 '18 at 10:20












          $begingroup$
          Given that OP has already found an upper bound by Cauchy-Schwarz, it suffices to find the values of $a$, $b$, $c$ whose sum attains that upper bound to finish the question. Anyways, it's pedagical to use Lagrange multipliers.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 11 '18 at 10:33




          $begingroup$
          Given that OP has already found an upper bound by Cauchy-Schwarz, it suffices to find the values of $a$, $b$, $c$ whose sum attains that upper bound to finish the question. Anyways, it's pedagical to use Lagrange multipliers.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 11 '18 at 10:33











          1












          $begingroup$

          With no existing correct answer, I feel I have to add one. Actually, that's a classic inequalities problem that doesn't even need calculus, and has a one-line solution.



          By $s$-power inequality, $4(a+b+c) le 4 cdot 3 cdot sqrt{dfrac{a^2+b^2+c^2}{3}} = 40 sqrt3$. Equality holds iff $a = b = c = sqrt{dfrac{10^2}{3}} = dfrac{10 sqrt3}{3}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, the am-gm inequality says that :$$frac{x_1+ ldots + x_n}{n} geq sqrt[n]{x_1 cdots x_n}$$. How does this gives the above result?
            $endgroup$
            – nafhgood
            Dec 11 '18 at 10:39










          • $begingroup$
            @mathnoob Sorry for the name confusion. It should be called the $s$-power inequality (called "generalized mean inequality" in Wiki. AM-GM is just a special case of that inequality.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:43
















          1












          $begingroup$

          With no existing correct answer, I feel I have to add one. Actually, that's a classic inequalities problem that doesn't even need calculus, and has a one-line solution.



          By $s$-power inequality, $4(a+b+c) le 4 cdot 3 cdot sqrt{dfrac{a^2+b^2+c^2}{3}} = 40 sqrt3$. Equality holds iff $a = b = c = sqrt{dfrac{10^2}{3}} = dfrac{10 sqrt3}{3}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, the am-gm inequality says that :$$frac{x_1+ ldots + x_n}{n} geq sqrt[n]{x_1 cdots x_n}$$. How does this gives the above result?
            $endgroup$
            – nafhgood
            Dec 11 '18 at 10:39










          • $begingroup$
            @mathnoob Sorry for the name confusion. It should be called the $s$-power inequality (called "generalized mean inequality" in Wiki. AM-GM is just a special case of that inequality.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:43














          1












          1








          1





          $begingroup$

          With no existing correct answer, I feel I have to add one. Actually, that's a classic inequalities problem that doesn't even need calculus, and has a one-line solution.



          By $s$-power inequality, $4(a+b+c) le 4 cdot 3 cdot sqrt{dfrac{a^2+b^2+c^2}{3}} = 40 sqrt3$. Equality holds iff $a = b = c = sqrt{dfrac{10^2}{3}} = dfrac{10 sqrt3}{3}$.






          share|cite|improve this answer











          $endgroup$



          With no existing correct answer, I feel I have to add one. Actually, that's a classic inequalities problem that doesn't even need calculus, and has a one-line solution.



          By $s$-power inequality, $4(a+b+c) le 4 cdot 3 cdot sqrt{dfrac{a^2+b^2+c^2}{3}} = 40 sqrt3$. Equality holds iff $a = b = c = sqrt{dfrac{10^2}{3}} = dfrac{10 sqrt3}{3}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 10:41

























          answered Dec 11 '18 at 10:27









          GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

          12.8k72445




          12.8k72445












          • $begingroup$
            Sorry, the am-gm inequality says that :$$frac{x_1+ ldots + x_n}{n} geq sqrt[n]{x_1 cdots x_n}$$. How does this gives the above result?
            $endgroup$
            – nafhgood
            Dec 11 '18 at 10:39










          • $begingroup$
            @mathnoob Sorry for the name confusion. It should be called the $s$-power inequality (called "generalized mean inequality" in Wiki. AM-GM is just a special case of that inequality.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:43


















          • $begingroup$
            Sorry, the am-gm inequality says that :$$frac{x_1+ ldots + x_n}{n} geq sqrt[n]{x_1 cdots x_n}$$. How does this gives the above result?
            $endgroup$
            – nafhgood
            Dec 11 '18 at 10:39










          • $begingroup$
            @mathnoob Sorry for the name confusion. It should be called the $s$-power inequality (called "generalized mean inequality" in Wiki. AM-GM is just a special case of that inequality.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 11 '18 at 10:43
















          $begingroup$
          Sorry, the am-gm inequality says that :$$frac{x_1+ ldots + x_n}{n} geq sqrt[n]{x_1 cdots x_n}$$. How does this gives the above result?
          $endgroup$
          – nafhgood
          Dec 11 '18 at 10:39




          $begingroup$
          Sorry, the am-gm inequality says that :$$frac{x_1+ ldots + x_n}{n} geq sqrt[n]{x_1 cdots x_n}$$. How does this gives the above result?
          $endgroup$
          – nafhgood
          Dec 11 '18 at 10:39












          $begingroup$
          @mathnoob Sorry for the name confusion. It should be called the $s$-power inequality (called "generalized mean inequality" in Wiki. AM-GM is just a special case of that inequality.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 11 '18 at 10:43




          $begingroup$
          @mathnoob Sorry for the name confusion. It should be called the $s$-power inequality (called "generalized mean inequality" in Wiki. AM-GM is just a special case of that inequality.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 11 '18 at 10:43


















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