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Wize High School Grade 12 Calculus Textbook > Derivative Applications

Optimization Problems

Maximum & Minimum on a Closed IntervalPrevious Section
Increasing & Decreasing IntervalsNext Section
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Optimization Problems

We want to find the maximum and minimum values in a real-world application problem.

Strategy
1. Determine what you are minimizing/maximizing and draw a picture if possible
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
3. If there is more than 1 independent variable, create another equation relating the variables → combine into one function with 1 independent variable
4. Follow the 3 steps to find the maximum or minimum value required
  • Step 1: find the derivative of the function and set to 0
  • Step 2: evaluate the function values at the point(s) found in step 1 and the endpoints of the interval (sometimes the endpoints don't exist)
  • Step 3: compare the values you found in step 2 and make a conclusion
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Example: Optimization

A farmer wishes to create a rectangular fenced enclosure for his animals. He has 600 m 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
See video for picture

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×wA=l\times w

We have 2 independent variables (lenght 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=6002l+w=600

3. Combine these to get an equation with an independent and a dependent variable
From the second equation, we have w=600−2lw=600-2l.
Substitute this into the area equation:
A=l(600−2l)A=l\left(600-2l\right)
A=600l−2l2A=600l-2l^2

4. Find the max value
Step 1: Find the derivative and set to 0
A′=600−4lA'=600-4l
0=600−4l0=600-4l
l=150l=150

Step 2: Evaluate the function at l=150 & w=300l=150\ \&\ w=300 and the endpoints l=0 & w=600l=0\ \&\ w=600 and l=300 & w=0l=300\ \&\ w=0
  • A=150×300=45000A=150\times300=45000
  • A=0×600=0A=0\times600=0
  • A=300×0=0A=300\times0=0
Therefore, the dimensions that will maximize the enclosure are l=150ml=150m and w=300mw=300m.
The maximum area is 150×300=45000m2150\times300=45000m^2.

Practice: Optimization

Four square corners are cut from a rectangular sheet measuring 160 cm by 160cm, and the remaining shape is bent to form an open-topped box. Find the side length of the cut-out square that maximizes the volume of the box.

Example: Optimization

Find two positive real numbers whose product is 1600, such that the sum of these two numbers is minimized.

1. No diagram required.
Let xx and yy be the two positive numbers.

2. We want to minimize the sum:
S=x+yS=x+y

3. We also know that xy=1600 → x=1600yxy=1600\ \to\ x=\frac{1600}{y}
So we get S=1600y+yS=\frac{1600}{y}+y

4. Minimize this function:
S′=−1600y2+1S'=-\frac{1600}{y^2}+1
0=−1600y2+10=-\frac{1600}{y^2}+1
1600y2=1\frac{1600}{y^2}=1
y2=1600y^2=1600
y=1600=40y=\sqrt{1600}=40 (we don't have to consider −40-40 since the real numbers are positive)

Find the sum at y=40y=40 and the endpoints (y=1, y=1600y=1,\ y=1600):
  • S=160040+40=80S=\frac{1600}{40}+40=80
  • S=16001+1=1601S=\frac{1600}{1}+1=1601
  • S=16001600+1600=1601S=\frac{1600}{1600}+1600=1601
Therefore, the two positive numbers are 40 and 4040\ and\ 40
Find the dimensions of the rectangle of largest area that has its base on the xx-axis, its other two vertices above the xx-axis and is laying on the parabola y=20−x2y=20-x^2.

Find the minimum vertical distance between the graphs of y1=x2+x+10y_1=x^2+x+10 and y2=−x2+x+2y_2=-x^2+x+2.
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Example: Optimization (Revenue)

A high school committee is selling tickets for a play. When the tickets are $7, they will set 2000 tickets. The Grade 12 math class did some research and found that for every $0.10 increase in ticket price, 40 fewer people will purchase a ticket. If the auditorium capacity for this play is 2600 people, and the play requires a minimum of 1600 people to perform, what price should the school committee set the ticket price at to maximize revenue while meeting these restrictions?

1. No diagram required.
Let xx represent the number of $0.10 increases.

2. We want to maximize revenue:
R(x)=(Number of tickets sold)×(Price of each ticket)R\left(x\right)=\left(\text{Number of tickets sold}\right)\times\left(\text{Price of each ticket}\right)
R(x)=(2000−40x)×(7+0.10x)R\left(x\right)=\left(2000-40x\right)\times\left(7+0.10x\right)
R(x)=14000−280x+200x−4x2R\left(x\right)=14000-280x+200x-4x^2
R(x)=14000−80x−4x2R\left(x\right)=14000-80x-4x^2

3. We have an equation that only has one independent variable.

4. Maximize this function:
R′(x)=−80−8xR'\left(x\right)=-80-8x
0=−80−8x0=-80-8x
x=−10x=-10

Evaluate the revenue at x=−10x=-10 and at the endpoints (x=−15x=-15 and x=10x=10):
  • R(−10)=14000−80(−10)−4(−10)2=14400R\left(-10\right)=14000-80\left(-10\right)-4\left(-10\right)^2=14400
  • R(−15)=14000−80(−15)−4(−15)2=14300R\left(-15\right)=14000-80\left(-15\right)-4\left(-15\right)^2=14300
  • R(10)=14000−80(10)−4(10)2=12800R\left(10\right)=14000-80\left(10\right)-4\left(10\right)^2=12800
Therefore, the committee should set the ticket price to 7+0.10×(−10)= $67+0.10\times\left(-10\right)=\ \$6 per ticket.
A metal cylinder without a top is made to contain 64π64\pi cm3 of glue. Determine the dimensions of the can that will minimize the amount of metal used.