1. More About Optimization
Section 2.6
More About Optimization

2. #17, p181
EX: Find the dimensions of the closed rectangular box with square base and volume 8000 cubic centimeters that can be constructed with the least amount of material
Solution: V = x2y S=2x2+4xy
Knowing that V=8000 as well, set x2y=8000 and solve for y to get y Substituting gives the function

3. Taking the first derivative of S gives us

4. Setting S’ = 0 and solving for x we find that there is a horizontal
tangent when

5. Finding the Minimum
To see that S has a minimum at x = 20, consider the plot to the right.
We see that S’ < 0 to the left and S’ > 0 to the right (why?), so S(20) must be the minimum.

6. Another Example
Ex: Find the maximum area that can be fenced into a rectangular pen with 100 feet of fence using a barn as one side.
Solution: A = xy (length times width)
L = 2x + y (only one y since one side is the barn)
Since L = 100 as well, we get 100 = 2x + y and solve for y to get y = 100 – 2x.
Substituting for y gives a function
A(x) = x(100 – 2x) = 100x – 2x2

7. Computing the derivative A’(x) = 100 – 4x and then setting this
equal to zero gives 100 – 4x = 0 so that x = 25.
Is A(25) the maximum area? Note that x = 25 so y = 100 – 2(25)
= 50 and our pen is not a square.
The maximum occurs because A00 = -4 and A is concave down.
Note that the minimum of A occurs at the endpoints of the domain [0,50]

8. The Closed Interval Max/Min Theorem
If f(x) is continuous on [a,b] and differentiable on (a,b), then its maximum and minimum on [a,b] occurs at either
Points where f 0(x) = 0 in [a,b] or (critical points)
f(a) or f(b) (endpoints)
Strategy: find points where f 0 = 0 and see if f is bigger/smaller there or at endpoints

9. Example So the max of f is f(10) = 100 and the min of f is f(6) = -100
Ex: Find the maximum and minimum values of f(x) = x3 – 9x2 on the interval [0,10].
Solution: Compute f 0(x) = 3x2 –18x = 0 and solve for x to get x = 0 and x = 6. We compare
Endpoints: f(0) = 0 and f(10) = 100
Crit Pts: f(0) = 0 and f(6) = -100
So the max of f is f(10) = 100 and the min of f is f(6) = -100
Graph this function to verify these facts

10. Revenue = (number sold)(price)
Another type of quantity that is of interest in business applications is revenue. Revenue is defined as
Revenue = (number sold)(price)
= x p(x)
If x = number sold and p(x) is a function that determines price as a function of x

11. Example
#1, p 195 An artist is planning to sell signed prints. If 50 prints are offered for sale she can charge $400 each. For each print she makes more than 50, she must lower the price by $5. How many prints should the artist make in order to maximize her revenue?
Solution: price = (beginning price) + (change in price)
= (5)(50 – x)
= – 5x
= 650 – 5x
And so revenue is R(x) = x ( 650 – 5x)
= 650x – 5x2
The max occurs when R0 (x) = 650 – 10x = 0 or when x = 65.

12. Economic Order Quantities
Another application. Want to minimize cost of keeping items in inventory.
See the Excel spreadsheet for more.