AP Calculus AB · Lesson 3

Optimization

Know Know it

Optimization problems ask for the largest or smallest value of a quantity. The method:

1. Draw and name the quantities. 2. Write the objective function, the quantity to optimize. 3. Use the constraint to write it in terms of one variable. 4. State the feasible domain: the inputs that make sense. 5. Find the critical points in that domain. 6. Justify the absolute extremum.

Two ways to justify an absolute extremum. On a closed interval, use the candidates test: compare the values at the critical points and the endpoints. On an open interval, if there is only one critical point and it is a relative extremum (shown by a sign change of f′f', or by the sign of f′′f'' there), that single relative extremum is the absolute one. Another way: if f′′>0f'' > 0 on the whole interval, the only critical point is the absolute minimum; if f′′<0f'' < 0 there, it is the absolute maximum.

Calculator On the exam

On the exam, a graphing calculator is required for the second part of the multiple-choice section (13 of the 42 questions, though not all of them need it) and for the first 2 of the 6 free-response questions. It is not allowed anywhere else. You may bring an approved handheld graphing calculator; the exam’s testing app also has a graphing calculator, available only in the calculator parts. Check the current calculator policy before test day.

Without a calculator, find critical points by factoring the derivative. With one, you can graph the derivative to see where it changes sign, or find a maximum value directly, but still write the derivative, the critical point and the reason in your answer.

Trap The best value can be at an endpoint

A critical point outside the feasible domain does not count, and an endpoint can win. If a profit function’s only critical point is at x=6x = 6 but the domain is 0≤x≤50 \leq x \leq 5, the maximum is at an endpoint. Always check the domain before you trust a critical point.

Worked example Easy

A lifeguard uses 80 meters of buoy rope to mark off a rectangular swimming area along a straight beach. The rope forms three sides of the rectangle, and the beach forms the fourth. What is the largest area the lifeguard can enclose?

  1. 20 square meters
  2. 400 square meters
  3. 800 square metersAnswer
  4. 1,600 square meters

How to solve it

  1. Let xx be each side that meets the beach. The rope gives 2x+y=802x + y = 80, so y=80−2xy = 80 - 2x.
  2. Area: A(x)=x(80−2x)=80x−2x2A(x) = x(80 - 2x) = 80x - 2x^2 on 0<x<400 < x < 40.
  3. A′(x)=80−4x=0A'(x) = 80 - 4x = 0 at x=20x = 20, and A′′(x)=−4<0A''(x) = -4 < 0, so this is the maximum: A(20)=800A(20) = 800 square meters.

Why each choice is right or wrong

  • A. Incorrect. 20 is the best width xx, in meters, the length of each side that meets the beach, not the area it gives.
  • B. Incorrect. 400 is the area of a 20 by 20 square, which uses rope on all four sides; the beach side needs no rope.
  • C. Correct. With width xx, the side along the water is 80−2x80 - 2x, so A=x(80−2x)A = x(80 - 2x). A′(x)=80−4x=0A'(x) = 80 - 4x = 0 at x=20x = 20, giving 20×40=80020 \times 40 = 800 square meters.
  • D. Incorrect. 1,600 is 40×4040 \times 40, which would need 120 meters of rope for three sides, more than the 80 meters available.

Explain Explain it

Task verbs: justify, determine

Write the justification in words. For example: "C′(x)<0C'(x) < 0 for 0<x<120 < x < 12 and C′(x)>0C'(x) > 0 for x>12x > 12, so CC decreases and then increases, and its only critical point, x=12x = 12, gives the absolute minimum."

Answer the question that was asked. If it asks for the dimensions, give both; if it asks for the least cost, give the cost, not the input.

Worked example Medium

A rectangular garden of area 200 square meters is fenced on all four sides. The fence along the side that faces the street costs 3 times as much per meter as the fence on the other three sides. What length of the street side gives the lowest total cost?

  1. 10 metersAnswer
  2. 10210\sqrt{2} meters
  3. 203\frac{20}{\sqrt{3}} meters
  4. 20 meters

How to solve it

  1. Constraint: xy=200xy = 200, so y=200xy = \frac{200}{x}. Cost (in units of the cheaper fence): C=3x+x+2y=4x+400xC = 3x + x + 2y = 4x + \frac{400}{x}, for x>0x > 0.
  2. C′(x)=4−400x2=0C'(x) = 4 - \frac{400}{x^2} = 0 gives x2=100x^2 = 100, so x=10x = 10.
  3. C′C' goes from negative to positive at x=10x = 10, the only critical point, so the cost is lowest there.

Why each choice is right or wrong

  • A. Correct. With street side xx and other side y=200xy = \frac{200}{x}, the cost is proportional to 3x+x+2y=4x+400x3x + x + 2y = 4x + \frac{400}{x}. Its derivative 4−400x24 - \frac{400}{x^2} is 0 at x=10x = 10.
  • B. Incorrect. 10210\sqrt{2} is the side of a 200-square-meter square, the best shape only if all four sides cost the same.
  • C. Incorrect. 203\frac{20}{\sqrt{3}} comes from 3x+400x3x + \frac{400}{x}, which leaves out the side opposite the street, whose fence costs 1 per meter.
  • D. Incorrect. 20 meters is the length of each of the other two sides, y=20010y = \frac{200}{10}, not the street side.

Justify Your turn

The cost, in dollars, of a delivery boat trip at a steady speed of vv kilometers per hour is C(v)=0.1v2+200vC(v) = 0.1v^2 + \frac{200}{v}, for v>0v > 0. Its derivative is C′(v)=0.2v−200v2C'(v) = 0.2v - \frac{200}{v^2}. Find the speed that makes C′(v)=0C'(v) = 0. Then use the sign of C′C' on each side of that speed to explain why it gives the lowest cost for every v>0v > 0.

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A full answer does this

  • Finds v=10v = 10 kilometers per hour from C′(v)=0C'(v) = 0
  • Says C′(v)<0C'(v) < 0 for 0<v<100 < v < 10 and C′(v)>0C'(v) > 0 for v>10v > 10
  • Concludes that CC decreases up to v=10v = 10 and increases after it, so v=10v = 10 gives the lowest cost for every v>0v > 0

Model answer

Setting C′(v)=0.2v−200v2=0C'(v) = 0.2v - \frac{200}{v^2} = 0 gives 0.2v3=2000.2v^3 = 200, so v3=1000v^3 = 1000 and v=10v = 10 kilometers per hour. For 0<v<100 < v < 10, C′(v)<0C'(v) < 0 (for example, C′(5)=−7C'(5) = -7), and for v>10v > 10, C′(v)>0C'(v) > 0 (for example, C′(20)=3.5C'(20) = 3.5). So CC decreases on 0<v<100 < v < 10 and increases for v>10v > 10, which means no speed gives a lower cost than v=10v = 10: the lowest cost is C(10)=30C(10) = 30 dollars.

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