Sam has sheets of metal which are 36 cm by 36 cm square. He wants to cut out identical squares which are x cm by I cm from the corners of each sheet. He will then bend the sheets along the dashed lines to form an open container. a Show that the volume of the container is given by V(2) = x(36 – 2x)2 cm? . b What sized squares should be cut out to produce the container of greatest capacity? 36 cm 36 cm

a) The volume of the box ([tex]V[/tex]), in cubic centimeters, is equal to the area of the base ([tex]A[/tex]), in square centimeters, multiplied by the height of the box ([tex]h[/tex]), in centimeters. The area of the surface is the product of the width ([tex]w[/tex]) and length (l), both in centimeters:

[tex]V = w\cdot l\cdot h[/tex]

[tex]V = (36-2\cdot x)^{2}\cdot x[/tex] (1)

The volume of the container is [tex]V = (36-2\cdot x)\cdot x[/tex] cubic centimeters. [tex]\blacksquare[/tex]

b) We need to apply first derivative analysis and second derivative analysis to determine the dimensions of the maximum squares to be cut:

FDT

[tex]-4\cdot (36-2\cdot x)\cdot x+(36-2\cdot x)^{2} = 0[/tex]

[tex]-4\cdot x + 36-2\cdot x = 0[/tex]

[tex]6\cdot x = 36[/tex]

[tex]x = 6[/tex]

SDT

[tex]V'' (x) = -4\cdot [(-2)\cdot x+(36-2\cdot x)]-4\cdot (36-2\cdot x)\cdot x + (36-2\cdot x)^{2}[/tex]

[tex]V''(x) = 8\cdot x-4\cdot (36-2\cdot x)-4\cdot (36\cdot x-2\cdot x^{2})+(36-2\cdot x)^{2}[/tex]

[tex]V'' (6) = -48[/tex]

Since [tex]V''(6) < 0[/tex], the critical value of [tex]x[/tex] leads to a maximum.

Four squares with size of 6 centimeters should be cut to produce the container of greatest capacity. [tex]\blacksquare[/tex]

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