optimization problem?

Question

a box with a square base and open top must have a volume of 32,000 cm^3. Find the dimension of the box that minimize the amount of material used.

Answer 1

If x = the length of a side of the square base and y = the height, then volume = lengthxwidthxheight

32000 = x^2y

y = 32000/x^2

There are 5 surfaces, 4 of which are rectangles with area xy and one square with area x^2

So surface area = 4xy + x^2

But y = 32000/x^2 so substitute to get

SA = 4x(32000/x^2) + x^2

That is 128000/x + x^2, or, 128000x^-1 + x^2

Find the derivative and set it = to zero.
SA' = -128000x^-2 + 2x = 0

2x = 128000/x^2

2x^3 = 128000

x^3 = 64000

Find the cube root of 64000; that's x

Plug in to get y

Answer 2

Such a box is defined by two parameters: A, the length of the sides of the base, and B, the height of the box (both measured in cm). The volume of this rectangular solid is given by the products of its three dimensions, which is A^2 * B. This volume is constrained to be 32,000, so given one parameter, we may solve for the other:

A^2 * B = 32,000

B = 32,000 / A^2

The total amount of material used is the area of the bottom plus the area of the 4 walls, which is:

V = A^2 + 4 * A * B.

...substituting for B, we get:

V = A^2 + 4 * A * (32,000 / A^2)

...multiplying through:

V = A^2 + 4 * (32,000 / A)

V = A^2 + 128,000 / A

...which is:

V = A^2 + 128,000 * A^-1

Phew!

I'll leave it to you to take the first derivative, and solve for zero to find the optimum. Remember to check that the derivative is negative to the left of the optima and positive to the right to ascertain that it is a minimum.

Answer 3

If x is a side of the square and y is the height of the box, you are being asked to minimize x^2 + 4xy (bottom plus four sides) where (x^2)y = 32,000 cm^3 (volume).

The answer is x = 40, y = 20. That minimizes the material used at 4800 cm^2.

Answer 4

I hope you've let x be the length of the side of the base, and h the height, and written down the formulas for the volume and the surface area (square + 4 rectangles, since it's open at the top). If not, do it now and then scroll down to see if I got them right.






















V = (x^2)*h
A = x^2 + 4hx
Now we're told that V = 32 000, so put that in the first formula, make h the subject, and then sub this expression for h in the second formula. Now you have A as a function of x only and can use your standard process of differentiate, put deriv = 0 etc to find the minimum value. Then again check below to see if I got it right (probably not, because I'm going to do it in my head)




















dA/dx = 2x - 128000x^(-2)
Zero when x = 40, Check this gives minimum.
h = 32000/1600 = 20

Therefore
x = 40, h = 20.

Answer 5

lol, you must use "Single Variable Calculus" by James Stewart. Look in the answer book at 11 and 13. They are similar. And do your own homework.