Optimization problems are similar to related rates problems. They are related to maximizing or minimizing the volume, area, or cost (most often one of these) of a given situation.
An easy example
Find the maximum product of two numbers whose sum is 50.
The first step is to set up the equations. let the first number be x and the second y. You must also find the feasible domain, the range of values that and work.
There are two equations, primary and secondary. The primary equation is x*y=M (for max.) The secondary is x+y=50.
Next, evaluate one variable in terms of the other. We know that x+y=50 so x=50-y. Plug this value back into the primary equation.
(50-y) *y=M or 50y-y^2=M
Now, take the derivative of this equation. Because M is a max value of the function, it is equal to zero. 50-2y=0. Solve the equation for y. y=25. Return to the secondary equation to find x (which is 25). Now you must prove that 25 is a max. This function is easily derived again, resulting in -2. Because the function is concave down, it is a maximum.
Examples
A farmer is building a pen next to a river acting as one side of the fence. Fencing costs $10/ft. What dimensions will give the farmer the largest area if the farmer has $420 to spend?
Find a rectangular solid with a square base with the smallest possible surface area given a volume of 1337in^3.
Find the greatest sum of two numbers whose product is 150.
An easy example
Find the maximum product of two numbers whose sum is 50.
The first step is to set up the equations. let the first number be x and the second y. You must also find the feasible domain, the range of values that and work.
There are two equations, primary and secondary. The primary equation is x*y=M (for max.) The secondary is x+y=50.
Next, evaluate one variable in terms of the other. We know that x+y=50 so x=50-y. Plug this value back into the primary equation.
(50-y) *y=M or 50y-y^2=M
Now, take the derivative of this equation. Because M is a max value of the function, it is equal to zero. 50-2y=0. Solve the equation for y. y=25. Return to the secondary equation to find x (which is 25). Now you must prove that 25 is a max. This function is easily derived again, resulting in -2. Because the function is concave down, it is a maximum.
Examples
A farmer is building a pen next to a river acting as one side of the fence. Fencing costs $10/ft. What dimensions will give the farmer the largest area if the farmer has $420 to spend?
Find a rectangular solid with a square base with the smallest possible surface area given a volume of 1337in^3.
Find the greatest sum of two numbers whose product is 150.