# Applied Maxima and Minima Problem

• Apr 14th 2010, 04:57 PM
Pebbles
Applied Maxima and Minima Problem
The base and top of a rectangular covered box are three times as long as they are wide. If the volume is a maximum for fixed total surface area, find the proportion of the height of the box to the short side of its base.

Does anyone know how to solve this??
• Apr 14th 2010, 05:38 PM
Let:
x = width of box
h = height of box
V = Volume of box
A = Surface Area of box
P = Proportion of the height of the box to the side of the base
then:
length of box = 3x

$V = 3x^2h$

$A = 2x.h+2.3x.x+2.3x.h = 8xh+6x^2=2x(4h+3x)$

$P = \frac{h}{x}$

Now for constant A we can make h a function of x by re-arranging formula for A as follows:
$\frac{A}{2x}=4h+3x$

$4h=\frac{A}{2x}-3x$

$h=\frac{A}{8x}-\frac{3}{4}x$

And so now we can calculate V as a function of x:
$V = 3x^2h=3x^2(\frac{A}{8x}-\frac{3}{4}x) = \frac{3A}{8}x-\frac{9}{4}x^3$

And to find the maximum V we need $\frac{dV}{dx}=0$ so:
$\frac{dV}{dx}=\frac{3A}{8}-\frac{27}{4}x^2=0$

$\frac{3A}{8}=\frac{27}{4}x^2$

$A=18x^2$

But we also know that $A = 2x(4h+3x)$ so:
$2x(4h+3x)=18x^2$

$4h+3x=9x$

$4h=6x$

$2h=3x$

$\frac{h}{x}=\frac{3}{2}$

So $P=\frac{3}{2}$

Which is the solution:
Proportion of the height of the box to the side of the base is 1.5 !
• Apr 14th 2010, 05:40 PM
palabine
It always helps to draw a picture. Let x be the short side and h be the height. Then the box is x by 3x by h so:

V = 3x^2*h. This is what you want to maximize, i.e. take derivative and find critical points. But you need to use surface area to solve and substitute for h.

Surface area is a known constant, let's call "SA". SA is the four sides plus the top and bottom. The top and bottom are 3x^2 each and there are 2 sides xh and 2 sides 3xh. So:

SA = 3x^2 + 3x^2 + 2xh + 6xh = 6x^2 + 8xh.

Solve for h = (SA-6x^2)/8x, substitute into V and take derivative of V with respect to x, solve for zero. That will be best x, called Xmax. Then substitute Xmax into above h equation to find Hmax and take the proportion.

Hope this helps.

[This question took 6 minutes to answer]
• Apr 14th 2010, 05:44 PM
skeeter
Quote:

Originally Posted by Pebbles
The base and top of a rectangular covered box are three times as long as they are wide. If the volume is a maximum for fixed total surface area, find the proportion of the height of the box to the short side of its base.

$L = 3W$

$V = LWH = 3W^2H$

$H = \frac{V}{3W^2}$

desired ratio is ...

$\frac{H}{W} = \frac{V}{3W^3}$

surface area ...

$A = 2(LW+WH+HL)$

$A = 2\left(3W^2 + \frac{V}{3W} + \frac{V}{W}\right)$

if $V$ is a max, then $\frac{dA}{dW} = 0$ ...

$\frac{dA}{dW} = 2\left(6W - \frac{V}{3W^2} - \frac{V}{W^2}\right) = 0$

$6W = \frac{4V}{3W^2}$

$W^3 = \frac{2V}{9}$

sub into the earlier expression for $\frac{H}{W}$ ...

$\frac{H}{W} = \frac{V}{3} \cdot \frac{9}{2V} = \frac{3}{2}$