Minimizing surface area for a given volume
$begingroup$
Math question:An open-top box with a square base is to have a volume of 4 cubic ft. Find the dimensions of the box that can be be made with the smallest amount of material.
This is the only thing I got: $V=x^2y=4$
calculus
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
$endgroup$
add a comment |
$begingroup$
Math question:An open-top box with a square base is to have a volume of 4 cubic ft. Find the dimensions of the box that can be be made with the smallest amount of material.
This is the only thing I got: $V=x^2y=4$
calculus
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
$endgroup$
1
$begingroup$
Ok, so the equation for $V$ is a constraint on your length variables. Now write down an expression for surface area in terms of $x$ and $y$.
$endgroup$
– Simon S
Apr 17 '15 at 23:11
add a comment |
$begingroup$
Math question:An open-top box with a square base is to have a volume of 4 cubic ft. Find the dimensions of the box that can be be made with the smallest amount of material.
This is the only thing I got: $V=x^2y=4$
calculus
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
$endgroup$
Math question:An open-top box with a square base is to have a volume of 4 cubic ft. Find the dimensions of the box that can be be made with the smallest amount of material.
This is the only thing I got: $V=x^2y=4$
calculus
calculus
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
edited Apr 17 '15 at 23:11
Simon S
23.2k63481
23.2k63481
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
asked Apr 17 '15 at 23:10
4everyoung194everyoung19
12
12
1
$begingroup$
Ok, so the equation for $V$ is a constraint on your length variables. Now write down an expression for surface area in terms of $x$ and $y$.
$endgroup$
– Simon S
Apr 17 '15 at 23:11
add a comment |
1
$begingroup$
Ok, so the equation for $V$ is a constraint on your length variables. Now write down an expression for surface area in terms of $x$ and $y$.
$endgroup$
– Simon S
Apr 17 '15 at 23:11
1
1
$begingroup$
Ok, so the equation for $V$ is a constraint on your length variables. Now write down an expression for surface area in terms of $x$ and $y$.
$endgroup$
– Simon S
Apr 17 '15 at 23:11
$begingroup$
Ok, so the equation for $V$ is a constraint on your length variables. Now write down an expression for surface area in terms of $x$ and $y$.
$endgroup$
– Simon S
Apr 17 '15 at 23:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So you have $V=xcdot x cdot y=4, $ i.e. your volume is $4$ and this equation has two variables. We need a second equation that way we will have $2$ equations with $2$ variables right?
So whats the surface area of an open top box? In general the it will be (using $w$=width for base, $l$=length for base, and $h$=height)
$S=wcdot l+2 cdot w cdot h+2cdot lcdot h $ (only one $w cdot l$ because its open top). Based on your question it looks like $y$ is the height and $w=l=x$.
Solve $V$ for one variable say $y=frac{4}{x^2}$. Put that into $S$ and calculate $frac{dS}{dy}=0$. Find the $y$ works and if there's only one its probably the min. If there's more than one you have to check to the right and left to see if its a min or max.
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
answered Nov 10 '15 at 23:24
RashadRashad
34
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add a comment |
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$begingroup$
So you have $V=xcdot x cdot y=4, $ i.e. your volume is $4$ and this equation has two variables. We need a second equation that way we will have $2$ equations with $2$ variables right?
So whats the surface area of an open top box? In general the it will be (using $w$=width for base, $l$=length for base, and $h$=height)
$S=wcdot l+2 cdot w cdot h+2cdot lcdot h $ (only one $w cdot l$ because its open top). Based on your question it looks like $y$ is the height and $w=l=x$.
Solve $V$ for one variable say $y=frac{4}{x^2}$. Put that into $S$ and calculate $frac{dS}{dy}=0$. Find the $y$ works and if there's only one its probably the min. If there's more than one you have to check to the right and left to see if its a min or max.
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
answered Nov 10 '15 at 23:24
RashadRashad
34
$endgroup$
add a comment |
$begingroup$
So you have $V=xcdot x cdot y=4, $ i.e. your volume is $4$ and this equation has two variables. We need a second equation that way we will have $2$ equations with $2$ variables right?
So whats the surface area of an open top box? In general the it will be (using $w$=width for base, $l$=length for base, and $h$=height)
$S=wcdot l+2 cdot w cdot h+2cdot lcdot h $ (only one $w cdot l$ because its open top). Based on your question it looks like $y$ is the height and $w=l=x$.
Solve $V$ for one variable say $y=frac{4}{x^2}$. Put that into $S$ and calculate $frac{dS}{dy}=0$. Find the $y$ works and if there's only one its probably the min. If there's more than one you have to check to the right and left to see if its a min or max.
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
answered Nov 10 '15 at 23:24
RashadRashad
34
$endgroup$
add a comment |
$begingroup$
So you have $V=xcdot x cdot y=4, $ i.e. your volume is $4$ and this equation has two variables. We need a second equation that way we will have $2$ equations with $2$ variables right?
So whats the surface area of an open top box? In general the it will be (using $w$=width for base, $l$=length for base, and $h$=height)
$S=wcdot l+2 cdot w cdot h+2cdot lcdot h $ (only one $w cdot l$ because its open top). Based on your question it looks like $y$ is the height and $w=l=x$.
Solve $V$ for one variable say $y=frac{4}{x^2}$. Put that into $S$ and calculate $frac{dS}{dy}=0$. Find the $y$ works and if there's only one its probably the min. If there's more than one you have to check to the right and left to see if its a min or max.
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
answered Nov 10 '15 at 23:24
RashadRashad
34
$endgroup$
So you have $V=xcdot x cdot y=4, $ i.e. your volume is $4$ and this equation has two variables. We need a second equation that way we will have $2$ equations with $2$ variables right?
So whats the surface area of an open top box? In general the it will be (using $w$=width for base, $l$=length for base, and $h$=height)
$S=wcdot l+2 cdot w cdot h+2cdot lcdot h $ (only one $w cdot l$ because its open top). Based on your question it looks like $y$ is the height and $w=l=x$.
Solve $V$ for one variable say $y=frac{4}{x^2}$. Put that into $S$ and calculate $frac{dS}{dy}=0$. Find the $y$ works and if there's only one its probably the min. If there's more than one you have to check to the right and left to see if its a min or max.
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
answered Nov 10 '15 at 23:24
RashadRashad
34
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
edited Nov 10 '15 at 23:43
daOnlyBG
2,31371734
2,31371734
answered Nov 10 '15 at 23:24
RashadRashad
34
answered Nov 10 '15 at 23:24
RashadRashad
34
answered Nov 10 '15 at 23:24
RashadRashad
34
34
add a comment |
add a comment |
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$begingroup$
Ok, so the equation for $V$ is a constraint on your length variables. Now write down an expression for surface area in terms of $x$ and $y$.
$endgroup$
– Simon S
Apr 17 '15 at 23:11