Unconstrained Optimization of One Variable

J. Zachary Klingensmith

6 Unconstrained Optimization of One Variable

6.1 the First derivative test

Practice Problems

Problem 6.1.1: Use the first-derivative test to find the relative extrema of the function below.

$y=5x^2-30x+6$

Answer: Relative min at x=3.

Problem 6.1.2 Use the first-derivative test to find the relative extrema of the function below.

$f(x)=x^{2/3}-4$

Answer: Relative min at x=0.

Problem 6.1.3: Use the first-derivative test to find the relative extrema of the function below.

$g(x)=\frac{x^2}{x^2-4}$

Answer: Relative max at x=0.

Problem 6.1.4: Use the first-derivative test to find the relative extrema of the function below.

$h(x)=(x-2)^3$

Answer: No relative extrema

Problem 6.1.5: Use the first-derivative test to find the relative extrema of the function below.

$f(x)=-2x^3+39x^2-240x+3$

Answer: Relative min at x=5, relative max at x=8

Problem 6.1.6: Use the first-derivative test to find the relative extrema of the function below.

$g(x)=7(x-5)^{2/3}$

Answer: Relative min at x=5.

Problem 6.1.7: The cost of manufacturing $Q$ toasters in one day is given by

$C(Q)=0.05Q^2+24Q+300, 0<Q<150$

a) Find the average total cost function.
b) Find the relative extrema of the ATC function.

Answer:

$ATC=.05Q+24+\frac{300}{Q}$

Minimized at Q=77.46.

Problem 6.1.8: For the following function, determine the interval(s) for which the function is increasing and the interval(s) for which the function is decreasing:

$y=5x^2-30x+6$

Answer:

Decreasing from $(-\infty,3)$ and increasing from $(3,\infty)$

Problem 6.1.9: For the following function, determine the interval(s) for which the function is increasing and the interval(s) for which the function is decreasing:

$f(x)=-2x^3+39x^2-240x+3$

Answer:

Decreasing from $(-\infty,5)\cup(8,\infty)$ and increasing from $(5,8)$ .

External Resources

Khan Academy: Introduction to Minimum and Maximum Points

Khan Academy: Finding Relative Extrema (First Derivative Test)

Khan Academy: Worked Example (First Derivative Test)

Khan Academy: Analyzing Mistakes – First Derivative Test

Khan Academy: Analyzing Mistakes II (First-Derivative Test)

7.2 higher-order derivatives and the second derivative test

Practice Problems

Problem 6.2.1: Find $g''(x)$ for the following function:

$g(x)=x^4-5x^2-1.$

Answer:

$g''(x)=12x^2-10$

Problem 6.2.2: Find the first three derivatives of the following function:

$y=7x^4+8x.$

Answer:

$y'=29x^3+8, y''=84x^2, y'''=168x$

Problem 6.2.3: Let $f(x)=(5-8x^2)^2$ . Find the first three derivatives and evaluate each at $x=3$ .

Answer:

$f'(x)=256x^3-80x, f'(3)=6672; f''(x)=768x^2-80, f''(3)=6832; f'''(x)=1536x, f'''(3)=4608$

Problem 6.2.4: Determine the point(s) of inflection of the following function:

$f(x)=4x^4+54x^3-42x^2+2.$

Answer: 1/4, 7

Problem 6.2.5: Determine the interval(s) the following function is concave up and concave down:

$f(x)=12x^5+45x^4-360x^3+5.$

Answer: Concave up from $(4.32,0)\cup(2.08,\infty)$ and concave down from $(-infty,-4,32)\cup(0,2.08)$

Problem 6.2.6: Determine the interval(s) the following function is concave up and concave down:

$g(x)=-x^4-2x^3+3x-8.$

Answer: Concave up from $(-1,0)$ and concave down from $(-\infty,-1)\cup(0,\infty)$ .

Problem 6.2.7: Find the relative extrema of the function below with the second derivative test. If the second derivative test is inconclusive for a critical value, use the first derivative test. Verify that what you found is indeed a max/min.

$y=4x^2-32x+6$

Answer: Relative min at x=4.

Problem 6.2.8: Find the relative extrema of the function below with the second derivative test. If the second derivative test is inconclusive for a critical value, use the first derivative test. Verify that what you found is indeed a max/min.

$f(x)=2x^3-3x^2-12x+5$

Answer: Relative max at x=-1, relative min at x=2

Problem 6.2.9: Find the relative extrema of the function below with the second derivative test. If the second derivative test is inconclusive for a critical value, use the first derivative test. Verify that what you found is indeed a max/min.

$g(x)=x+\frac{4}{x}$

Answer: Relative max at x=-1, relative min at x=1

Problem 6.2.10: Find the relative extrema of the function below with the second derivative test. If the second derivative test is inconclusive for a critical value, use the first derivative test. Verify that what you found is indeed a max/min.

$h(x)=-2x^3+39x^2-216x+11$

Answer: Relative max at x=9, relative min at x=4

Problem 6.2.11: Find the relative extrema of the function below with the second derivative test. If the second derivative test is inconclusive for a critical value, use the first derivative test. Verify that what you found is indeed a max/min.

$f(x)=12x^5+60x^4-240x^3+7$

Answer: Relative max at x=-6, relative min at x=2

Problem 6.2.12: Find the relative extrema of the function below with the second derivative test. If the second derivative test is inconclusive for a critical value, use the first derivative test. Verify that what you found is indeed a max/min.

$g(x)=(x+2)(x-3)^2$

Answer: Relative max at x=1/3, relative min at x=3.

Problem 6.2.13: The moose population on an island is found using the function below. At what time is the moose population maximized? What is the maximum population? Verify that what you found is indeed a maximum.

$P(t)=120t-0.4t^4+1200$

Answer: 4.2 years; max population is 1580

Problem 6.2.14: The revenue from selling $Q$ items is $R(Q)=400Q-Q^2$ and the total cost is given by $C(Q)=175+7Q$ . Find the quantity which optimizes profit. Show that the quantity produces a maximum. What is the maximum profit? Verify your results.

Answer: Max profit at Q=196.5, max profit is 38,437.25.

Problem 6.2.15: A manufacturer has determined that the total cost $C$ (in dollars) of operating a factory is $C(Q)=0.5Q^2+10Q+7200$ where $Q$ is the number of units produced. Further, average total cost is calculated by dividing the total cost by $Q$ . Find the quantity which minimizes the average total cost of production.