Question

A box with an open to is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. find the largest volume that such a box can have?

Answer 1

`V= 2ft^3`

Explanation:

Step 1: get equation for volume
`V= x(3-2x)^2`

Step 2: Differentiate using chain rule and product rule
`V'=1*(3-2x)^2 + x(2)(3-2x)(-2)`
`V'=9-12x+4x^2-12x+8x^2`
`V'= 9-24x+12x^2`

Step 3: Find the critical numbers by find where V'=0 or V' DNE
`V' = 0= 9-24x+12x^2`
(note: V' DNE does not apply in this problem)

Step 4: factor to solve
`V' = 0= 9-24x+12x^2`
`0= 3(4x^2-6x+3)`
`0= (2x-3)(2x-1)`

`x=3/2, 1/2`

Step 5: rule out `x=3/2` because it would cause the sides of the box to be negative. Plug `x=1/2` into original volume equation.
`V(1/2)=2`