From a rectangular 5×13 piece of cardboard, four congruent squares with sides of length x are cut out, one at each corner. (See the accompanying figure.) The resulting crosslike piece is then folded to form a box. Answer the following.
a) Find the volume V of the box as a function of x.
b) Sketch the graph of y=V(x).

The volume of the cuboid is the measure of the space occupied within a cuboid. The cuboid is a three-dimensional shape that has length, breadth, and height.

a)Each side of the future box is

L = 5-2x ,D = 13-2x

and let the height of box is x

Then, volume of the box as function of x is:

V(x)= (5-2x)*(13-2x)*x

V(x)= (65-10x-26x +4[tex]x^{2}[/tex])*x

V(x)= 65x-36[tex]x^{2}[/tex]+4[tex]x^{3}[/tex]

Differentiate the above equation, we get

V'(x)= 65-72x+12[tex]x^{2}[/tex]

If V'(x) = 0, then

65-72x+12[tex]x^{2}[/tex]=0

[tex]x_{1,2} = (-b\pm \sqrt{ b^{2} - 4*a*c} ] /2a[/tex]

[tex]x_1[/tex]=4.89297

[tex]x_2[/tex]=1.10703

We will take x= 1.10703

Then the sides of the box are:

L = 5-2(1.10703) = 2.21406 ft

D = 13 - 2*x = 13 - 2x(1.10703)= 10.78594 ft

H = x = 1.10703 ft

Now, volume is

V = 2.21406*10.78594*1.10703 = 24.4366 ft³

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