User:Cehjohnson

Problem
A rectangular pizza is placed on a tray. The sides of the tray extend by 4 units to the right and left and by 3 units (x) above and below. Calculate a value of x such that the area of the tray protruding is as small as possible.



Answer
The solution to this problem lies in differential calculus. Looking at the diagram below, we can see that the area of the tray extending from the pizza is given by



Substituting:

$${\mathrm y} = \frac{60}x$$

$$A = (\frac{60}x + 6)(x + 8) - 60$$

$$A = \frac{60x}x + 6x + \frac{480}{x} - 60$$

$$A = 6x + \frac{480}{x}$$





$$\frac{{\mathrm d}A}{{\mathrm d}x} = 6 - \frac{480}{x^2}$$

Since the area should be at a minimum:

$$0 = 6 - \frac{480}{x^2}$$

$$x = \sqrt80$$