At Domino’s, they deliver for you. Their square…


At Domino’s, they deliver for you. Their square pizza boxes are formed from 20 cm by 40 cm rectangular pieces of cardboard by cutting out six squares of equal size, three from each of the 40 cm edges (one cut at each comer and one from the middle of the 40 cm edge). What’s left is folded in the obvious fashion. How does one obtain the box of largest volume?

Finding Pizza Box of Maximum Volume

Square pizza boxes for Domino’s are made from rectangular pieces of cardboard of given dimension. To make these pizza boxes, equal-sized square pieces of cardboard are cut from the four corners and the two middles of the longer side. Then what’s left is folder along the shorter side to get us the pizza box. Using Geometry, Algebra and Calculus we find the size of the squares we need to cut out so that we get a pizza box of maximum volume. The first derivative and second derivative test from Calculus will be used to find us the value of the size of the squares.

Answer and Explanation:

First, six squares having side x cm each are cut out from the four corners and in the middle of the 40 cm side. The the cardboard is folder along the 20 cm side to get a pizza box.

The dimensions of the pizza box are length L = (20 – 2x) cm, width W = (20 – x – x/2) = (20 – 3x/2) cm and height H = x cm.

Then the volume of the box will be

{eq}V = LWH = (20-2x)(20-3x/2)(x) = 3x^3-70x^2+400x qquad (1) {/eq}

To obtain the box of maximum volume, we need to maximize the volume function in (1). To that end we find its critical points by calculating the first derivative as follows, from Calculus:

{eq}V'(x) = 9x^2-140x+400 qquad (2) {/eq}

Since V'(x) is defined everywhere, we find critical points by solving V'(x) = 0 to get

{eq}9x^2-140x+400=0 qquad (3) {/eq}

Using the quadratic equation, we solve (3) to get

{eq}displaystyle x = frac {140 pm sqrt {(-140)^2-4(9)(400)}}{2(9)} = 3.77, ; 11.78. {/eq}

So we have two critical points. To determine which one gives us a box of maximum volume we find the second derivative of V to get

{eq}V”(x) = 18x-140 qquad (4) {/eq}

Using (4) we get that {eq}V”(3.77) < 0 ; {rm and} ; V”(11.78)>0. {/eq} So, from the second derivative test, x = 3.77 is a point of maximum for V(x).

The six squares will need to be of size 3.77 cm each to get a box of maximum volume.

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