calculus! problem

Question

An open box is to be made out of a 6-inch by 20-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.

Dimensions of the bottom of the box:  

Height of the box

Answer 1

Height = x

Sides are (6-2x) and (20-2x)

Volume, V(x) =x(6-2x)(20-2x) = (6x-2x²)(20-2x)

Find V'(x), set to zero and find x

V'(x)=(6-4x)(20-2x)+(6x-2x²)(-2) = 120-12x-80x+8x²-12x+4x² = 12x²-92x+120 = 0

x=1.371, 7.296

V(1.371) ≈ 77.09

V(7.296) ≈-339.01

Choose x = 1.371 because it produces the largest volume (or you can use the 2nd derivative test)

Height = 1.371 inches

Side1 = (6-2(1.371)) = 3.258 inches

Side2 = (20-2(1.371)) = 17.258 inches

Second derivative test:

V''(x) = 24x-104

V''(1.371) = -71.1 < 0 --> Maximum

V''(7.296) = 71.104 >0 --> Minimum

Answer 2

Let's assume the rectangle is oriented with the 20 inch side along a horizonal axis and the 6 inch side along the vertical.

We'll be cutting a square of side length x from each of the four corners of the rectangle. Once folded, this will give an open box whose length and width will both be reduced by 2*x, and whose height will be x.

So the dimensions of the open box will be 20 - 2x, 6 - 2x, and x. (Note that this constrains x to be < 3 for a valid solution since otherwise the removal of the square corner will completely remove the side of length 6.)

With these dimensions, the volume V of the box is length * width * height =

(20 - 2x)(6 - 2x)(x) =

(20 - 2x)(6x - 2x²) =

120x - 12x² - 40x² + 4x^{3 =} 120x - 52x² + 4x³

To find the min or max values for this expression, take the first derivative:

dV/dx = 120 - 104x + 12x²

Setting this expression to 0 and using the quadratic formula yields two roots: 7.30 and 1.37. (approx)

The first value can be eliminated because it's larger than the 3" constraint above.

To confirm that the second value is in fact a maximum, plug it into the second derivative:

d²V/dx2 = 24x - 104

Evaluating at x=1.37 yields -71. The fact that this is negative confirms that we had found a max value.