A box is formed by cutting squares from the four corners of a sheet of paper and folding up the sides.

Question

The graph below shows how the volume of the box in cubic inches, V, is related to the length of the side of the square cutout in inches, x.

The graph of the function V equals x times the quantity 7 minus two x times the quantity 9 minus two x. The point 1.3 comma 36.61 is indicated on the graph and represents the cutout length that produces the maximum volume of the box. V(1.3, 36.61)

Suppose the largest possible cutout length is 3.5 inches. Over what interval of x does the volume of the box decrease as the cutout length gets larger? (Enter your answer as an interval.)

Answer 1

V = x(7 - 2x)(9-2x), where 0 < x < 3.5

V = x(63 - 32x + 4x²) = 4x³ - 32x² + 63x

V' = 12x² - 64x + 63 = 0

x = [64 ± √1072] / 24 = 4.03, 1.30

Since x < 3.5, we can delete 4.03. So, V' = 0 when x = 1.30

When 0 < x < 1.3, V' > 0 . So, V is increasing.

When 1.3 < x < 3.5, V' < 0. So, V is decreasing.

Answer 2

Here is what the graph looks like:

desmos.com/calculator/mmuj8m9msa

When x is in the interval (1.3, 3.5) the volume of the box is decreasing.

The inference is that the rectangular sheet is 7'' by 9". By cutting x" squares from each corner, then folding up the sides. the dimensions of the box are-- width 7 - 2x, length: 9 - 2x, and height: x (the side length of the square cutout). That is why the volume function is V(x) = LWH = x(7 - 2x)(9 - 2x). I used calculus to determine the exact value that gives the maximum volume--but all you are concerned about is where is the volume decreasing (where does the graph go "downhill" as you move from left to right along the x-axis).