Calculus and Maximum Volume?

Question

1. Take a piece of paper 8 inches by 10 inches and cut squares of the same size
out of each of the four corners (see next page).
2. Let x denote the length and width of each corner square.
3. Fold up the four sides to form an open box.
4. Write the volume as a function of x.
5. Determine the size of the squares that will yield the box with the largest
volume.
6. Determine the maximum volume.

Answer 1

i will not answer the question for you, this is meant for you to learn, but here is a general way to solve optimization problems:
1) write down an expression for the entity you are trying to optimize. together with domain.
2) if the function has more than one variable, write as a function of one variable using known relations.
3) set the derivative = 0 and solve.
4) verfiy that the answer obltained gives optimum value.
for the volume of a box formed out of a sheet which has dimensions a by b, the volume is
V= x(a-2x)(b-2x) 0 solve v'=0
check if solution gives max volume.

Answer 2

Let x = the side of each square
Then x will be the height of the box
10 - 2x will be the length of the box
8-2x will be the width of the box
So volume = lwh (10-2x)(8-2x)x
V= 4x^3 - 36x^2 +80x
dV/dx = 12x^2 -72x +80
Set above to 0 to find critcal points
x = [72 +/- 8sqrt(21)]/24
x = 3 +/- sqrt(21)/3
x = 3 +/- 1.527
x = 1.472
x = 4.527
The max occurs at x = 1.472 and the min at x = 4.527
The size of the squares are 1.472 per side
the max volume is 1.472(10-2*1.472)(8-2*1.472)
= 1.472*7.056*5.056 = 52.51 in^3