MODULE 6
Calculus Based Optimization
TEACHING SUGGESTIONS
Teaching Suggestion M6.1: Why Discuss Calculus?
Many of the quantitative models are derived with the help of calculus. The economic order quan-
Teaching Suggestion M6.2: Graphical Presentation Aids in Understanding Maxima and Minima.
When discussing the maximum and minimum values for a function, it is helpful to draw a graph
SOLUTIONS TO DISCUSSION QUESTIONS
M6-2. To find the slope of a nonlinear function at a particular point, we find the slope of a line
M6-3. To find the maximum or minimum of a function, we take the first derivative of this func-
M6-4. A critical point is a point where the first derivative equals zero.
SOLUTIONS TO PROBLEMS
M6-5. a. Y = 2(3)X3–1 – 3(2)X2–1 + 0 = 6X2 – 6X
b. Y = 4(5)X5–1 + 2(3)X3–1 – 12(1)X1–1 = 20X4 + 6X2 – 12
M6-6. a. Y = 12X – 6
b. Y = 80X3 + 12X
M6-7. a. Y = 6X5 – 0.5(2)X = 6X5 – X
b. Y = 20X3 + 24X + 10
M6-8. a. Y = 30X4 – 1
b. Y = 60X2 + 24
M6-9. Y = 12X – 5. Set this equal to 0 and solve.
12X – 5 = 0
X = 5/12
M6-10. Y = X2 – 10X + 25. Set this equal to 0 and solve.
M6-11. Y = 3X2. Set this equal to 0.
M6-12. TR = 1,200 – 0.5Q
1,200 – 0.5Q = 0
M6-13. Total revenue = TR = QP
TR = (75 – 2P)P
TR = 75P – 2P2
To maximize total revenue, find the first derivative of TR.
M6-14. Total Revenue TR = QP = (180 – 2P2)P
TR = 180P – 2P3
M6-15.
( )
2,000 25 10 2,000 40
2
Q
TC Q
= + +
50,000 5 80,000TC Q
Q
= + +
M6-16.
( )
2 1 3
50,000 2 100,000
TC QQ
+
−−
==