Optimization: Other Applications

Transcription

1 Optimization: Other Applications MATH 151 Calculus for Management J. Robert Buchanan Department of Mathematics Fall 2018

2 Objectives After completing this section, we will be able to: use the concepts of maximum and minimum in applications related to geometry and science, and use the concepts of maximum and minimum in applications using the Pythagorean Theorem, volume, area, distance, and velocity. We now turn to one of the most important applications of the derivative: finding the maximum or minimum of a function.

3 Optimization on Closed Intervals A continuous function f on a closed interval [a, b] has both an absolute maximum and an absolute minimum. To find them: 1. Find all the critical numbers for f in [a, b]. 2. Evaluate f at the critical numbers and at the endpoints a and b. The largest and smallest values found in Step 2 will be the absolute maximum and minimum values of f on [a, b].

4 Guidelines Draw a picture, if applicable. Determine the variables of the problem and how they are related. Decide which quantity should be optimized. Find an expression for the quantity to be optimized. This expression should contain only one variable. Determine the minimum and maximum allowable values (if any) for the variable. Apply the First or Second Derivative Test to optimize the quantity.

5 Useful, Common Formulas Pythagorean Theorem : a 2 + b 2 = c 2 Area of triangle : A = 1 2 b h Area of rectangle : A = l w Circumference of circle : C = 2πr Area of circle : A = πr 2 Surface area of cube : S = 6x 2 Volume of a box : V = l w h Volume of sphere : V = 4 3 πr 3 Volume of cone : V = 1 3 πr 2 h

6 Example Farmer MacDonald has 300 feet of chicken wire with which to construct a rectangular pen to hold a flock of chickens. A 400-ft long chicken coop will be used to form one side of the pen, so the wire is needed for the remaining three sides. How can the pen be constructed so that the birds have the maximum space in which to roam?

7 Solution Let y be the width of the pen and x be the length of the pen. 300 = x + 2y A = xy = (300 2y)y = 300y 2y 2 da dy = 300 4y 0 = 300 4y = y = 75 (critical number) A(0) = 0 A(75) = 11, 250 (local maximum) A(150) = 0

8 Example: Volume Equal squares are to be cut from the corners of a rectangular piece of sheet metal that has dimensions 10 inches by 16 inches. Find the dimensions of the box having the maximum volume.

9 Solution Let the squares cut from the corners be x by x inches with 0 x 5. V = x(10 2x)(16 2x) = 4x 3 52x x dv dx = 12x 2 104x = 4(3x 20)(x 2) 0 = x 2 = x = 2 (critical number) V (0) = 0 V (2) = 144 (local maximum) V (5) = 0

10 Example: Pipeline Construction An oil company wishes to run a pipeline from a drilling platform located 5 miles offshore to a shipping terminal 16 miles down the coast. The costs are $130, 000 per mile to lay the pipeline underwater and $120, 000 per mile to lay the pipeline over land. Find the location on the coast between the drilling platform and the shipping terminal that will minimize the total cost of the pipeline.

11 Solution Let x be the distance down the coast that the underwater pipeline comes ashore with 0 x 16. Using the Pythagorean Theorem we see that 16 x = length of overland pipeline x = length of underwater pipeline C(x) = (x ) 1/ (16 x) C (x) = x x = x = x = 12 (critical numbe x C(0) = 2, 570, 000 C(12) = 2, 170, 000 (local minimum) C(16) = 2, 179, 200

12 Example: Distance and Velocity A projectile is fired vertically, and its height (in feet) after t seconds is given by s(t) = 104t 16t Find the maximum height of the projectile. 2. When does the projectile hit the ground? 3. How fast is the projectile moving when its hits the ground?

13 Solution 1. Maximum height: s(t) = 104t 16t 2 s (t) = t 0 = t = t = 3.25 (critical number) s (t) = 32 < 0 s(3.25) = 169 (maximum height) 2. Ground strike time 0 = 104t 16t 2 = t(104 16t) t = 0 or t = Ground strike velocity s (7.5) = (7.5) = 136

14 Example (1 of 2) The value of a timber forest t years after it is planted is given by the function V (t) = 96 t 6t (units of thousands of dollars). Find the year when the value of the forest is maximized.

15 Example (1 of 2) The value of a timber forest t years after it is planted is given by the function V (t) = 96 t 6t (units of thousands of dollars). Find the year when the value of the forest is maximized. In this example we have 0 t <. V (t) = 96t 1/2 6t V (t) = 48t 1/2 6

16 Example (1 of 2) The value of a timber forest t years after it is planted is given by the function V (t) = 96 t 6t (units of thousands of dollars). Find the year when the value of the forest is maximized. In this example we have 0 t <. V (t) = 96t 1/2 6t V (t) = 48t 1/2 6 To find the critical numbers we must solve the equation: 48t 1/2 6 = 0 8t 1/2 = 1 t 1/2 = 8 t = 64

17 Example (2 of 2) Since there is only one critical number, we may use the Second Derivative Test to determine if it is a maximum or minimum point.

18 Example (2 of 2) Since there is only one critical number, we may use the Second Derivative Test to determine if it is a maximum or minimum point. V (t) = 24t 3/2 = 24 t 3/2 V (64) = 24 = 1.5 < 0 642/3 Thus the value of the forest is maximized 64 years after planting. The value is V (64) = (64) = 384 thousand dollars.

19 Marginal Analysis Recall that profit equals revenue minus cost. P(x) = R(x) C(x) Profit is maximized at a critical point. P (x) = R (x) C (x) 0 = R (x) C (x) R (x) = C (x) Profit is maximized when marginal revenue equals marginal cost.