Optimization ( Applied Max and Minimum )

[asz304]

Homework Statement

From a square piece of cardboard, 30 cm on each side, an open topped box is to be constructed by cutting the squares from the corners and turning up the sides. What are the dimensions of the box of largest volume?

The Attempt at a Solution

I know how to do the derivatives from the equation below. but from the book I don't get why the base of the box is : 30 - 2x.

The equation goes like this:
V(x) = x(30-2x)^2.

 

[Dest]

base of the box would be 30 - 2x because you are subtracting x twice. Try to draw a diagram to help you

so like you said

v(x) = x(30-2x)(30-2x)
v(x) = 4x^3 - 120x^2 + 900x
v`(x) = 12x^2 - 240x + 900
v`(x) = 12(x-15)(x-5)
v`(x) = 5, 15

Therefore dimestions of largest volume are 5, 20, 20

[note the interval was 0 < x < 15 (if you went higher than 15 you would have negative distance, you can close the intervals or open them depending on how you look at it- but in this case 15 would yield a minimum rather than a maximum)

Check:
v``(x) = 24x - 240
v``(5) = 24(5) < 240
so this is a maximum because v``(x) < 0