Calculus/Optimization

Introduction

Optimization is the use of Calculus in the real world. Let us assume we are a pizza parlor and wish to maximize profit. Perhaps we have a flat piece of cardboard and we need to make a box with the greatest volume. How does one go about this process?

Obviously, this requires the use of maximums and minimums. We know that we find maximums and minimums via derivatives. Therefore, one can conclude that Calculus will be a useful tool for maximizing or minimizing (also known as "Optimizing") a situation.

Examples

Example 1:

A box manufacturer desires to create a box with a surface area of 100 inches squared. 
What is the maximum size volume that can be formed by bending this material into a box? 
The box is to be closed.  The box is to have a square base, square top, and rectangular sides.

• Write out known formulas and information

$A_{base}=x^2$

$A_{side}=x\cdot h$

$A_{total}=2x^2+4x\cdot h=100$

$V=l\cdot w\cdot h=x^2\cdot h$

• Eliminate the variable h in the volume equation

$2x^2+4xh=100$

$x^2+2xh=50$

$2xh=50-x^2$

$h=\frac{50-x^2}{2x}$

$V$	=	$(x^2)\left(\frac{50-x^2}{2x}\right)$
	=	$\frac{1}{2}(50x-x^3)$

• Find the derivative of the volume equation in order to maximize the volume

$\frac{dV}{dx}=\frac{1}{2}(50-3x^2)$

• Set $\frac{dV}{dx}=0$ and solve for $x$

$\frac{1}{2}(50-3x^2)=0$

$50-3x^2=0$

$3x^2=50$

$x=\pm \frac{\sqrt{50}}{\sqrt{3}}$

• Plug-in the x value into the volume equation and simplify

$V$	=	$\frac{1}{2}[50\cdot \sqrt{\frac{50}{3}}-{\left(\sqrt{\frac{50}{3}}\right)}^3]$
	=	$68.04138174..$

Answer: $V_{max}=68.04138174..$

Sales Example

A small retailer can sell n units of a product for a revenue of r(n)=8.1n and at a cost of c(n)=n³-7n²+18n, with all amounts in thousands. How many units does it sell to maximize its profit?

The retailer's profit is defined by the equation p(n)=r(n) - c(n), which is the revenue generated less the cost. The question asks for the maximum amount of profit which is the maximum of the above equation. As previously discussed, the maxima and minima of a graph are found when the slope of said graph is equal to zero. To find the slope one finds the derivative of p(n). By using the subtraction rule p'(n)=r'(n) - c'(n):

$p(n)$	= $r(n)-c(n)$
$p^{\prime }(n)$	= $\frac{d}{dn}[8.1n]-\frac{d}{dn}[n^3-7n^2+18n]$
	= $-3n^2+14n-9.9$

Therefore, when $-3n^2+14n-9.9=0$ the profit will be maximized or minimized. Use the quadratic formula to find the roots, giving {3.798,0.869}. To find which of these is the maximum and minimum the function can be tested:

p(0.869) = - 3.97321, p(3.798) = 8.58802

Because we only consider the functions for all n >= 0 (i.e. you can't have n=-5 units), the only points that can be minima or maxima are those two listed above. To show that 3.798 is in fact a maximum (and that the function doesn't remain constant past this point) check if the sign of p'(n) changes at this point. It does, and for n greater than 3.798 P'(n) the value will remain decreasing. Finally, this shows that for this retailer selling 3,798 units would return a profit of $8,588.02.

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