Question

An open-top box is to be made from a 42-inch by 48-inch piece of plastic by removing a square from each corner of the plastic and folding up the flaps on each side. What should be the length of the side x of the square cut out of each corner to get a box with the maximum volume

Answer 1

Answer:

x  = 7.45 inch for the maximum volume.

The area of the square = x² = 7.45² = 55.5025  inch²

Step-by-step explanation:

From the given information:

An open-top box is to be made from a 42-inch by 48-inch piece of plastic by removing a square from each corner of the plastic and folding up the flaps on each side.

The objective is to determine the length of the side x of the square cut out of each corner to get a box with the maximum volume

The volume of the box = l×b×h

The volume of the box = $(42 - 2x) \times (48-2x) \times (x)$

The volume of the box = $(2016 - 84x - 96x +4x^2)x$

The volume of the box = $(2016 -180x+4x^2)x$

The volume of the box = $(2016x -180x^2+4x^3)$

The volume of the box = $4x^3 - 180x^2 +2016x$

For the maximum volume V' = 0

V' = $12x^2 - 360x + 2016$

Using the quadratic formula; we have:

$= \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$

where;

a = 12 , b = -360 c = 2016

$= \dfrac{-(-360) \pm \sqrt{(-360)^2 -4(12)(2016)}}{2(12)}$

$= \dfrac{360 \pm \sqrt{129600 -96768}}{24}$

$= \dfrac{360 \pm \sqrt{32832}}{24}$

$= \dfrac{360 \pm 181.196}{24}$

$= \dfrac{360 + 181.196}{24} \ \ \ OR \ \ \ \dfrac{360 - 181.196}{24}$

$= \dfrac{541.196}{24} \ \ \ OR \ \ \ \dfrac{178.804}{24}$

$= 22.55 \ \ \ OR \ \ \ 7.45$

For the maximum value , we check the points in the second derivative term

V'' = 24x - 360

V'' ( 22.55) = 24(22.55) - 360

V'' ( 22.55) = 541.2 - 360  

V'' ( 22.55) = 181.2   (minimum)

V'' ( 7.45) = 24(7.45) - 360

V'' ( 7.45) = 178.8 - 360  

V'' ( 7.45) = -181.2  < 0  (maximum)

Therefore, x  = 7.45 inch for the maximum volume.

The area of the square = x² = 7.45² = 55.5025  inch²

Answer 2

The maximum volume of a box is the highest volume the box can take.

The side length that ensures maximum volume is 22.55 inches or 7.45 inches.

The dimension of the plastic is:

$\mathbf{Length = 42}$

$\mathbf{Width = 48}$

Assume the side length cut-out is x

So, the dimension of the box is:

$\mathbf{Length = 42 - 2x}$

$\mathbf{Width = 48 - 2x}$

$\mathbf{Height = x}$

So, the volume of the box is:

$\mathbf{Volume = Length \times Width \times Height}$

This gives;

$\mathbf{Volume = (42 - 2x) \times (48 - 2x) \times x}$

Expand

$\mathbf{Volume = (42 - 2x) \times (48x - 2x^2)}$

Expand

$\mathbf{Volume = 2016x - 96x^2 - 84x^2 + 4x^3}$

Differentiate

$\mathbf{V' = 2016 - 192x - 168x + 12x^2}$

$\mathbf{V' = 2016 -360x + 12x^2}$

Rewrite as:

$\mathbf{V' = 12x^2 -360x + 2016}$

Set to 0

$\mathbf{12x^2 -360x + 2016 = 0}$

Divide through by 12

$\mathbf{x^2 -30x + 168 = 0}$

Using a calculator, the values of x are:

$\mathbf{x = 22.55\ or\ x = 7.45}$ ------ approximated to 2 decimal places

Hence, the side length that ensures maximum volume is 22.55 inches or 7.45 inches.

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