A rectangular sheet of cardboard 3 feet by 5 feet will be made into an open box by cutting equal-sized squares from each corner and folding up the four edges. A diagram of the unfolded box is provided in figure, the shaded region represents the area
that was cut from the corners of the box. What is the largest volume of such a box?

A rectangular sheet of cardboard 3 feet by 5 feet will be made into an open box by cutting equal-sized squares from each corner and folding up the four edges.

The shaded region represents the area that was cut from the corners of the box.

We have to determine,

The largest volume of such a box.

According to the question,

The height will be x

The width will be (3 − 2 x)

And the length will be ( 5− 2 x)

.

The formula for volume will be,

[tex]Volume = Length \times Height \times Width\\\\[/tex]

Substitute the values in the formula,

[tex]Volume = x(3-2x).(5-2x)\\\\Volume = (3x-6x^{2}).(5-2x)\\\\Volume = 3x(5-2x) - 6x^{2}(5-2x)\\\\Volume = 15x - 6x^{2} - 30x^{2} + 12x^{3}\\\\Volume = 12x^{3} -36x^{2} + 15x[/tex]

Then,

To Find the derivative of the equation,

[tex]V = 12x^{3} -36x^{2} + 15x\\\\Differentiate\ with\ respect\ to\ x\ both\ sides,\\\\\dfrac{dv}{dx} = \dfrac{d(12x^{3} -36x^{2} + 15x)}{dx}\\\\V' = 36x^{2} - 72x+15[/tex]

The critical numbers by seeing where V ' = 0,

[tex]= 36x^{2} -72x +15\\\\= 3 ( 12x^{2} - 24x +5 )\\[/tex]

By using the quadratic formula or your calculator to solve for

x = 0.23, and 1.6

x = 1.6 is not in the domain since that would make one of our side lengths less than zero. x = 0.23

Plug x back into the equations to get the dimensions,

[tex]3-2x = 3- 2 \times -0.23 = 3+0.46 = 3.46\\5-2x = 5 - 2 \times -0.23 = 5+0.46 = 5.46[/tex]

Therefore, the dimension are 0.23 ft x 3.46 ft x 5.46 ft.

Hence, The largest volume of such a box is 4.345[tex]ft^{3}[/tex].

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