Question 370483
V=(13-2h)(9-2h)(h)
Great work so far. To find the h that makes a box of maximum volume we will be finding the derivative of your volume function. (If you are not taking Calculus or you have never heard of a derivative, then you might as well stop reading and repost your question. If you do repost you might want to specify "without using Calculus" so other tutors don't give you Calculus-based solutions as well.)<br>
In general, maximum and/or minimum values for a function are found at any of the following:<ul><li>At the endpoints of the domain. For your volume function h must be between 0 and 4.5. We can't have negative sides to the box. If h < 0, then the height is negative. If h >4.5, then the 9-2h side will be negative. If h=0 then the volume is zero. And if h=4.5 then the 9-2h side is zero making the volume zero again. So the endpoints of your domain tell you minimum values for the volume.</li><li>Where the first derivative is either zero or undefined. So we need to find the first derivative:
The derivative of your volume function will be a little easier if we go ahead and multiply it out first:
{{{V = (117-44h+4h^2)(h)}}}
{{{V = 117h-44h^2+4h^3}}}
Now we can find the first derivative:
V' = {{{117-88h+12h^2}}}
Now we will use the derivative to find relative maximum and/or minimum points. These will occur where the first derivative is zero or undefined. This derivative is defined for all possible values of h. But we can find where it is zero:
{{{0 = 117-88h+12h^2}}}
This will not factor so we will need to use the Quadratic Formula:
{{{h = (-(-88) +- sqrt((-88)^2 -4(12)(117)))/2(12)}}}
which simplifies as follows:
{{{h = (-(-88) +- sqrt(7744 -4(12)(117)))/2(12)}}}
{{{h = (-(-88) +- sqrt(7744 - 5616))/2(12)}}}
{{{h = (-(-88) +- sqrt(2128))/2(12)}}}
{{{h = (88 +- sqrt(2128))/24}}}
{{{h = (88 +- sqrt(16*133))/24}}}
{{{h = (88 +- sqrt(16)*sqrt(133))/24}}}
{{{h = (88 +- 4*sqrt(133))/24}}}
{{{h = (4(22 +- sqrt(133)))/(4*6)}}}
{{{h = (cross(4)(22 +- sqrt(133)))/(cross(4)*6)}}}
{{{h = (22 +- sqrt(133))/6}}}
In long form this is:
{{{h = (22 + sqrt(133))/6}}} or {{{h = (22 - sqrt(133))/6}}}
If you get out your calculator you will find that the first h is greater than 4.5. This is not in the domain so we reject it. The other value for h, {{{(22 - sqrt(133))/6}}} then is the one which must create the box of maximum volume. This is true because<ul><li>The endpoints of the domain create minimum volume boxes.</li><li>The first derivative is defined for all values of h so no maximum or minimum values come from an undefined first derivative.</li><li>Therefore, the only h in the domain that makes the first derivative zero, {{{(22 - sqrt(133))/6}}}, is our answer because a maximum must exist at one of these three places.</li></ul></li></ul>