Optimization problem!?

Question

An open -top box is to be made by cutting small congruent squares from the corners of a 12-by12-in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible?

Answer 1

Hi,

If you cut a square x inches long from each corner, the width and length of the box will be 12 - 2x and the height of the sides bet up is x inches. So the volume = LWH is:

Y = x(12 - 2x)(12 - 2x)

If we look for a maximum volume where 0 < x < 6, since you can't cut more than 6 inches from each side because the whole side is only 12 inches. The maximum occurs when x = 2 and gives a volume of 128 cubic inches.

I hope that helps!! :-)

Answer 2

Both sides of the box will measure 12 - 2x while the height of the box will be x, where x is the measure of a side of a square to be cut from the corner.

Calculate volume:
(12 - 2x)(12 - 2x)(x)
(144 - 24x - 24x + 4x^2)(x)
144x - 48x^2 + 4x^3

Max volume will occur when the derivative of the function is equal to zero.

Take the first derivative:
144x - 48x^2 + 4x^3
144 - 96x + 12x^2

Solve the resulting quadratic for zero
144 - 96x + 12x^2 = 0
x = 6, x = 2

x cannot equal 6 because the sides of the sheet measures 12 in. If x = 6 then there would be no length and hence no volume.

Thus x must equal 2 and we should cut 2 in. by 2 in. squares from each corner to optimize the maximum volume.

Answer 3

Imagine the squares cut out of the four corners and assign them a value x. The height of the box, then, is x, and each side of the base is 12-2x.

V = x(12-2x)(12-2x) = 4x^3 - 48x^2 + 144x

Take the derivative of the function and find where its zeros (relative maximums and minimums)

V' = 12x^2 - 96x + 144
V' = 0
x = 2, x = 6

Since x = 0 and x = 6 are the minimum and maximum values for the cut-out squares you can reason that x = 2 is the correct answer, providing the largest volume. If you're not sure, you then make a table of values of x vs. V, making sure to include the endpoint values of x.

Answer 4

The height of the box will be x, and the length and width 12 - 2 x. The volume is

... V = x * (12 - 2 x)^2

The maximum is when the derivative is zero, that is,

... dV/dx = 0
... (12 - 2x)^2 - 4 x (12 - 2x) = 0
... (12 - 2x) (12 - 2x - 4x) = 0
... (12 - 2x) (12 - 6x) = 0
x = 6 or x = 2.

The first solution will give volume zero, and does not count. The second solution is the one you're looking for, with
... V = 2 x 8 x 8 = 128 sq in.

Answer 5

Hello

Cut x from each side then bend up the sides to form a box.

We have each side is 12- 2x and the depth is x so the volume is

(12-2x)(12-2x)x = v

or 144x - 48x^2 + 4x^3 = v

take the derivative and we get

144 -96x + 12 x^2 = V'

setting this equal to 0 and dividing by 12 give us

12 -8x + x^2 = 0

so (x-6)(x-2) = 0
x=2 or x = 6, but x can't be 6 or we have no sides so x = 2.

Hope this helps.

Answer 6

let x = width of square to be cut

V = (12-2x)*(12-2x)*x

V = x*(4x^2 - 48x + 144)

V = 4x^3 - 48x^2 + 144x

V' = 12x^2 - 96x + 144 = 0

x^2 - 8x + 12 = 0

(x - 6)(x - 2) = 0

x = 2 in, discard x = 6 in