# What is the maximum volume of the box, given the parameters below?

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Given a rectangular sheet of cardboard, #16# in. by #10# in., you are asked to cut off identical squares from each of the four corners of the sheet and then bend up the sides of the remaining cardboard to form a rectangular box.

Given a rectangular sheet of cardboard,

##### 1 Answer

The associated max volume is given by:

# V=144# cubic inches

#### Explanation:

Let us set up the following variables:

# {(w, "Width of the Box (in)"), (l, "Length of the Box (in)"), (x, "Length of the Corner Cut-out (in)"), (V, "Volume of the Box (cubic in)") :} #

We want to vary the corner length

The dimensions of the sheet are 16" by 10", hence, wlog (without loss of generality) taking

Width:

#x+w+x=16 \ \ => w=16-2x = 2(8-x)#

Length:#x+l+x=10 => l=10-2x = 2(5-x)#

Then the volume is given by:

# \ \ \ \ \ V=wlx#

# :. V = 2(8-x)2(5-x)x #

# :. V = 4x(x^2-13x+40)#

# :. V = 4x^3-52x^2+160x#

Differentiating wrt

# :. (dV)/dx=12x^2-104x+160 #

At a critical point,

# :. 12x^2-104x+160 = 0 #

# :. 3x^2-26x+40 = 0 #

# :. (3x-20)(x-2) = 0 #

# :. x=2,20/3 = 0 #

We should check while value leads to a maximum volume

# :. (d^2V)/dx^2=24x-104 #

# x = 2 => (d^2V)/dx^2 < 0 => max#

# x = 20/3 => (d^2V)/dx^2 > 0 => min#

So we have a maximum volume when

# V = 4*8-52*4+160*2=32-208+320=144#

If we graph the Volume Function,

graph{4x^3-52x^2+160x [-3, 10, -70, 160]}

Hopefully you can visually confirm the above #