1
Optimization Problems
In Optimization problems we are searching for a global maximum or minimum
of a function.
Steps to solving an Optimization problem
1. Read the problem carefully (read it 2 or 3 times if necessary). Identify what
quantity you are being asked to maximize or minimize. What are the unknowns in
the problem and what information are you given about them?
2. Draw a picture (if possible)
3. Assign letters to the unknown quantities and the thing you are trying to
maximize or minimize, Q (Q is called the Objective Function).
4. Express Q in terms of the quantities in the problem.
5. If necessary, find relationships in the problem that allow you to write Q in
terms of a single unknown. Write down the domain of Q.
6. Find the absolute maximum or minimum of Q.
7. Make sure you have answered the question raised in the problem.
Recall the following theorem:
Theorem (This is very useful!!) If 𝑓(𝑥) is a continuous function and 𝑐 is the only
critical point for 𝑓(𝑥) on an interval (open, closed, bounded or unbounded):
a. If 𝑐 is a relative minimum then, 𝑐 is an absolute minimum on that interval.
b. If 𝑐 is a relative maximum then, 𝑐 is an absolute maximum on that interval.
, 2
Ex. A farmer has 2400 ft. of fencing and wants to fence off a rectangular field that
borders a straight river. He needs no fence along the river. What are the
dimensions of the field that has the largest area?
2𝑊 + 𝐿 = 2400
𝐴 = 𝐿𝑊 (𝐴 is the objective function). 𝑊
We need 𝐴 to be a function of 1 variable
not 2 (𝐿, 𝑊).
𝐿
2𝑊 + 𝐿 = 2400 ⟹ 𝐿 = 2400 − 2𝑊
𝐴 = 𝐿𝑊 = 𝑊(2400 − 2𝑊); 0 ≤ 𝑊 ≤ 1200, since 0 ≤ 𝐿.
We want to find the global maximum of 𝐴 subject to 0 ≤ 𝑊 ≤ 1200.
2
𝐴 = 𝑊(2400 − 2𝑊) = 2400𝑊 − 2𝑊
𝐴′ (𝑊 ) = 2400 − 4𝑊 = 0 ⟹ 𝑊 = 600.
Determine whether 𝑊 = 600 is a local maximum/minimum or neither.
By the second derivative test:
𝐴′′ (𝑊 ) = −4 ⟹ 𝐴′′ (600) = −4 < 0 ⟹ 𝑊 = 600 a local max.
Since 𝑊 = 600 is the only critical point on 0 ≤ 𝑊 ≤ 1200, it's a global max.
𝑊 = 600 ⟹ 𝐿 = 2400 − 2(600) = 1200.
Dimensions of largest area: 𝑊 = 600𝑓𝑡, 𝐿 = 1200𝑓𝑡.
Optimization Problems
In Optimization problems we are searching for a global maximum or minimum
of a function.
Steps to solving an Optimization problem
1. Read the problem carefully (read it 2 or 3 times if necessary). Identify what
quantity you are being asked to maximize or minimize. What are the unknowns in
the problem and what information are you given about them?
2. Draw a picture (if possible)
3. Assign letters to the unknown quantities and the thing you are trying to
maximize or minimize, Q (Q is called the Objective Function).
4. Express Q in terms of the quantities in the problem.
5. If necessary, find relationships in the problem that allow you to write Q in
terms of a single unknown. Write down the domain of Q.
6. Find the absolute maximum or minimum of Q.
7. Make sure you have answered the question raised in the problem.
Recall the following theorem:
Theorem (This is very useful!!) If 𝑓(𝑥) is a continuous function and 𝑐 is the only
critical point for 𝑓(𝑥) on an interval (open, closed, bounded or unbounded):
a. If 𝑐 is a relative minimum then, 𝑐 is an absolute minimum on that interval.
b. If 𝑐 is a relative maximum then, 𝑐 is an absolute maximum on that interval.
, 2
Ex. A farmer has 2400 ft. of fencing and wants to fence off a rectangular field that
borders a straight river. He needs no fence along the river. What are the
dimensions of the field that has the largest area?
2𝑊 + 𝐿 = 2400
𝐴 = 𝐿𝑊 (𝐴 is the objective function). 𝑊
We need 𝐴 to be a function of 1 variable
not 2 (𝐿, 𝑊).
𝐿
2𝑊 + 𝐿 = 2400 ⟹ 𝐿 = 2400 − 2𝑊
𝐴 = 𝐿𝑊 = 𝑊(2400 − 2𝑊); 0 ≤ 𝑊 ≤ 1200, since 0 ≤ 𝐿.
We want to find the global maximum of 𝐴 subject to 0 ≤ 𝑊 ≤ 1200.
2
𝐴 = 𝑊(2400 − 2𝑊) = 2400𝑊 − 2𝑊
𝐴′ (𝑊 ) = 2400 − 4𝑊 = 0 ⟹ 𝑊 = 600.
Determine whether 𝑊 = 600 is a local maximum/minimum or neither.
By the second derivative test:
𝐴′′ (𝑊 ) = −4 ⟹ 𝐴′′ (600) = −4 < 0 ⟹ 𝑊 = 600 a local max.
Since 𝑊 = 600 is the only critical point on 0 ≤ 𝑊 ≤ 1200, it's a global max.
𝑊 = 600 ⟹ 𝐿 = 2400 − 2(600) = 1200.
Dimensions of largest area: 𝑊 = 600𝑓𝑡, 𝐿 = 1200𝑓𝑡.
