# Thread: Applied Maxima and Minima Problem

1. ## 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??

2. 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 !

3. 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]

4. 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 $\frac{H}{W}$ ...

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