Calculus question (might be easier)?

Question

trouble with this bc problem

find the maximum volume of a box that can be made by cutting out squares from the corners of an 8-inch by 15-inch rectangular sheet of cardboard and folding up the sides. Justify your answer.

how do you start this problem

Answer 1

Volume is width x height x length

If you call the size of the squares X, then the width of the box will

be  8 - 2X (since a square comes off each side)

Similarly, the length will be 15 - 2X

The height when folded up will be X

so V = (8 - 2X)(15-2X)X

Multiply this out, then take the derviative and find out where

it is 0.

Answer 2

perimeter of 8by15 = 46

you know that the perimieter will then equal 2(8-2x) +2(15-2x)

base bottom area will be  ((8-2x))((15-2x))

height will be x  

so volume = x(2(8-2x))(2(15-2x))

=x(8-2x)(15-2x)

=2x(4-x)(15-2x)

=2x(60-8x-15+2x^2)

=2x(2x^2 - 8x +45)

=4x^3 -16x^2+ 90

now differentiate and set = 0

find values of x, one will be the right answer

{12x^2 -32x = 0

3x^2-8x=0

x(3x-8)=0

i believe the highest value for the volume is when x = 8/3