According to postal regulations, a carton is classified as "oversized" if the sum of its height and girth (the perimeter of its base) exceeds 98 in. Find the dimensions of a carton with square base that is not oversized and has maximum volume.

V (max) = 8712.07 in³

Dimensions:

x (side of the square base) = 16.33 in

girth = 65.32 in

height = 32.67 in

Step-by-step explanation:

Let

x = side of the square base

h = the height of the postal

Then according to problem statement we have:

girth = 4*x (perimeter of the base)

and

4 * x + h = 98 (at the most) so h = 98 - 4x (1)

Then

V = x²*h

V = x² * (98 - 4x)

V (x) = 98*x² - 4x³

Taking dervatives (both menbers of the equation we have:

V' (x) = 196 x - 12 x² ⇒ V' (x) = 0

196x - 12x² = 0 first root of the equation x = 0

Then 196 - 12x = 0 12x = 196 x = 196/12

x = 16,33 in ⇒ girth = 4 * (16.33) ⇒ girth = 65.32 in

and from equation (1)

y = 98 - 4x ⇒ y = 98 - 4 (16,33)

y = 32.67 in

and maximun volume of a carton V is

V (max) = (16,33) ² * 32,67

V (max) = 8712.07 in³