# Calculus

A box (with no top) is to be constructed from a piece of cardboard of sides A and B by cutting out squares of length h from the corners and folding up the sides. Find the value of h that maximizes the volume of the box if
A = 7 and B = 12

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1. Draw a rectangle 7 x 12.

In the corners draw four squares of dimensions h x h.

It doesn't matter what the length of h is.

Dimensions of a new rectangle will be ( 7 - 2 h ) x ( 12 - 2 h )

The volume of the box will be:

Area of a new rectangle ∙ length h

V = ( 7 - 2 h ) ∙ ( 12 - h ) ∙ h

V = ( 84 - 24 h - 7 h + 2 h² ) ∙ h

V = ( 2 h² - 31 h + 84 ) ∙ h

V = 2 h³ - 31 h² + 84 h

The function has an exstrem (maximum or minimum) where the first derivative is zero.

In this case you need to find:

dV / dh = V′( h ) = 0

( 2 h³ - 31 h² + 84 h )′ = 2 ∙ 3 h² - 31 ∙ 2 h + 84 = 0

6 h² - 62 h + 84 = 0

The solutios are:

h = ( 31 -√457 ) / 6 ≈ 1.60374

and

h = ( 31 +√457 ) / 6 ≈ 8.7296

Now you have to second derivative test.

If V"( h ) < 0 then function has maximum at h.

If V" ( h ) > 0 then function has minimum at h.

V" ( h ) = 2 ∙ 6 h - 62

V" ( h ) = 12 h - 62

For

h = 1.60374

V" ( h ) = 12 ∙ 1.60374 - 62 = − 42​.75512 < 0

For

h = 8.7296

V" ( h ) = 12 ∙ 8.7296 - 62 = 42.7552 > 0

So

For h = ( 31 -√457 ) / 6 ≈ 1.60374 volume has a maximum.

For h = ( 31 +√457 ) / 6 ≈ 8.7296 volume has a minimum.

By the way dimension h = 8.7296 are impossible because for h = 8.7296 dimension 7 - 2 h would have a negative value and the length cannot be negative.

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