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

-1, -2, -3, -4, -6

step-by-step explanation:

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry