Optimization Problems

The process of finding an efficient way to carry out a task is the basis of an optimization problem.   Efficient could mean minimum cost, maximum profit, least time consuming, etc.   Optimization problems will always break down into two parts:

Constraint Equation: Relates the variables in the problem with a condition that has to be met.
Objective Function: The quantity that is trying to be maximized (or minimized) subject to the constraint.

Example

A rancher has \(400\) feet of fence for constructing a rectangular corral.   One side of the corral will be formed by the barn and requires no fencing at all.   Three exterior fences and two interior fences partition the corral into three rectangular horse pins.   Find the dimensions of the corral that will maximize the enclosed area.   What is the area of that corral?

Draw a picture to help: barn
Due to the fencing, the constraint equation is: \(y+4x=400\)
Want to maximize the area, so the objective function is: \(A=xy\)
Solve the contraint equation for \(y\) and substitue into the objective function to have only one variable: \(\begin{array}{l} y=400-4x \\ \implies A=x(400-4x) \\ \implies A=400x-4x^2\end{array}\)
Find the critical points and establish where a local maximum exists: \(\displaystyle{\begin{array}{l} \frac{dA}{dx}=400-8x \\ \implies 0=400-8x \\ \implies 50=x\end{array}}\)
Sign chart and First Derivative Test:

\(\begin{array}{c} + & 0 & - \\ \hline & x=50 & \end{array}\)

First Derivative Test indicates \(x=50\) is a local maximum.
Plug in to find the dimension for \(y\): \(y+4(50)=400\implies y=400-200\implies y=200\)
The dimensions of the corral will be \(200\)ft by \(50\)ft.
The area of the enclosed space is:   \(A=400(50)-4(50)^2=10000\)   square feet.