
Optimization
The technique of finding a function's maximum and minimums can be applied to Optimization problems. These problems involve find the optimal value of a function, usually with a constraint equation.
Procedure for Optimization Problems
- Determine what you are minimizing/maximizing and draw a picture if possible
- Write an equation for the quantity you are trying to maximize or minimize (If there is more than 1 independent variable, create another equation relating the variables)
- Combine these to get an equation with an independent and a dependent variable
- Find and classify the critical points of this final combined equation
- Answer the question

Example: Optimization
A farmer wishes to create a rectangular fenced enclosure for his animals. He has 600 ft. of fencing material and will construct the enclosure using a pre-existing fence as one side. Find the dimensions of the fenced area to maximize the enclosure.
1. Determine what you are minimizing/maximizing and draw a picture if possible
We want to MAXIMIZE the area of the rectangle
2. Write an equation for the quantity you are trying to maximize or minimize
*If there is more than 1 independent variable, create another equation relating the variables
The equation for area is A=l×w
We have 2 independent variables (length and width), so we need another equation that relates these 2 variables. Since the fence only has to make up 3 of the 4 sides of the rectangle, the equation is: 2l+w=600
3. Combine these to get an equation with an independent and a dependent variable
From the second equation, we have w=600−2l.
Substitute this into the area equation:
A=l(600−2l)
A=600l−2l2
4. Find and classify the critical points of this final combined equation
A′=600−4l and A′′=−4
To find the critical points, we set the first derivative = 0:
600−4l=0 → l=150
Check to make sure this is a maximum:
A′′(150)=−4<0
So, this critical point corresponds to a maximum.
5. Answer the question
Therefore, the dimensions that will maximize the enclosure are l=150 ft and w=300 ft.
The maximum area is 150×300=45000 ft2.

Example: Optimization
Find the point on the graph of the equation y=x,x≥0, that is the closest to the point (3,0).
1. Determine what you are minimizing/maximizing and draw a picture if possible
We are trying to minimize the distance between y=x and (3,0).
2. Write an equation for the quantity you are trying to maximize or minimize
*If there is more than 1 independent variable, create another equation relating the variables
We want to optimize the distance formula d=(x−x0)2+(y−y0)2
However, for purposes of the derivative, it is equivalent and easier to use the square of the distance formula.
d2=(x−x0)2+(y−y0)2
Since this equation has more than 1 independent variable, substitute the constraint equation y=x
3. Combine these to get an equation with an independent and a dependent variable
d2=(x−x0)2+(x−y0)2
Also we may substitute (3,0) for x0 and y0
d2=(x−3)2+(x−0)2
=(x−3)2+x
4. Find and classify the critical points of this final combined equation
dxd[d2]=dxd(x−3)2+x
⇒0=2(x−3)+1
⇒x−3=−21
⇒x=25→critical point
f′′(x)=2>0∴ local, min, at x=25
5. Answer the question
Substitute x=25 into y=x⟹y=25
So the point (25,25) on x is closest to (3,0)