Please Help ASAP

Question

An open box is to be made from a 16 inch by 6 inch piece of cardboard by cutting squares of equal size from the four corners and bending up the sides. What size should the squares be to obtain a box with the largest possible volume?

I got 8.297 or -0.96418

Answer 1

x 16-2x x

|--------|-------------------------------|--------|

x | | | | x

|--------| |--------|

| |

| | 6 - 2x

| |

|--------| |--------|

x | | | | x

|--------|-------------------------------|--------|

x 16-2x x

Let the length of each side of each square be x.

The base of the open box is (16-2x) by (6-2x) and the height of the box is x.

Therefore, the volume of the box is V = (16-2x) * (6-2x) * x = (96 - 32x - 12x + 4x²) * x = 4x³ - 44x² + 96x

For the largest volume, we have,

dV/dx = 0

12x² - 88x + 96 = 0

3x² - 22x + 24 = 0

3x² - 22x + 24 = 0

3x² - 18x - 4x + 24 = 0

3x(x - 6) - 4(x - 6) = 0

(3x - 4) (x - 6) = 0

Therefore, x = 6 or 4/3

Now, for x = 6, d²V/dx² becomes [24x - 88]_x=6 = 56 > 0

And, for x = 4/3, d²V/dx² becomes [24x - 88]_x=4/3 = -56 < 0

Therefore, V is maximum for x = 4/3

Answer 2

Let x be the squares you cut out. The volume of the box will consist of sides length x and a base of (16-2x) by (6-2x)

V = wlh = (6-2x)(16-2x)x = 4x³ - 44x² + 96x

In order to find critical points, set dV/dx = 0

12x² - 88x +96 = 0

which has the roots (Quad Form) 4/3 and 6, so 4/3 is the only valid critical point. We know that the extremes (x =0 and x=3) yield a volume of 0, so we know that this point is a max for a polynomial. You can check for a + to - change in the derivative around the critical point or you can check the second derivative:

24x - 88 < 0 negative --> frown, i.e, maximum (positive would be smile, minimum)

x = 4/3 inch and V = (6-8/3)(16-8/3)(4/3) = 59.26 in³

Take care.