AP®MathsCollege BoardCalculus ABRevision NotesUnit 5: Analytical Applications of DifferentiationGraphs of Functions & Their DerivativesOptimization Problems

Optimization Problems (College Board AP® Calculus AB)

Revision Note

Jamie Wood

Expertise

Maths

Optimization problems

What is an optimization problem?

  • Recall that differentiation is about the rate of change of a function and provides a way of finding minimum and maximum values of a function

  • Anything that involves maximizing or minimizing a quantity can be modelled using differentiation; for example

    • minimizing the cost of raw materials used in manufacturing a product

    • finding the maximum height a football reaches when kicked

  • These are called optimization problems

How do I solve an optimization problem?

  • In optimization problems, variables other than x, y and f are often used

    • V is often used for volume, S for surface area

    • r for radius if a circle, cylinder or sphere is involved

  • Derivatives can still be found

    • but be clear about which letter is representing the independent (x) variable

    • and which letter is representing the dependent (y) variable

  • Problems often start by linking two connected quantities together, for example volume and surface area

    • If more than one variable is involved, constraints will usually be given

      • so that the quantity being optimized can be rewritten in terms of one variable

  • Once the quantity being optimized is written as a function of a single variable, differentiation can be used to maximize or minimize the quantity as required

  • STEP 1

    Rewrite the quantity to be optimized in terms of a single variable, using any constraints given in the question

  • STEP 2

    Differentiate and solve the derivative equal to zero to find the “x"-coordinate(s) of any critical points

  • STEP 3

    If there is more than one critical point, or you are required to justify the nature of the critical point, differentiate again

  • STEP 4

    Use the second derivative to determine the nature of each critical point and select the maximum or minimum point as necessary

  • STEP 5

    Interpret the answer in the context of the question

Worked Example

A large flower bed is being designed as a rectangle with a semicircle on each end, as shown in the diagram below.

The total area of the bed is to bespace 100 pi meters squared.

Diagram of an oval with two straight sides and semi-circular ends. A vertical line marks radius "r cm" from the center to the edge of a semi-circle.

(a) Show that the perimeter of the bed is given by the formula

space P equals pi stretchy left parenthesis r plus 100 over r stretchy right parenthesis

Answer: 

The width of the rectangle is 2 r meters, and its length is L meters

The area consists of a semi circle, plus a rectangle, plus another semi circle

1 half pi r squared space plus space 2 r L space plus thin space 1 half pi r squared space equals space 100 pi

Simplify and write L in terms of r

table row cell pi r squared plus 2 r L end cell equals cell 100 pi end cell row cell 2 r L end cell equals cell 100 pi minus pi r squared end cell row L equals cell fraction numerator 50 pi over denominator r end fraction minus pi over 2 r end cell end table

The perimeter of the flower bed consists of two semi-circular arcs, and two straight lengths

P equals pi r plus pi r plus 2 L

Substitute in the expression for L, and simplify to the desired expression for P

table row P equals cell pi r plus pi r plus 2 open parentheses fraction numerator 50 pi over denominator r end fraction minus pi over 2 r close parentheses end cell row P equals cell 2 pi r plus fraction numerator 100 pi over denominator straight r end fraction minus pi r end cell row P equals cell pi r plus fraction numerator 100 pi over denominator r end fraction end cell end table

space P equals pi stretchy left parenthesis r plus 100 over r stretchy right parenthesis

(b) Find the value of r that minimizes the perimeter, and find the minimum value of the perimeter.

Answer:

To find the minimum value of P, we need to find the minimum point on a graph of P against r

To do this, we need to differentiate P with respect to r

Start by writing P in terms of r, using the answer from part (a)

table row cell space P end cell equals cell pi stretchy left parenthesis r plus 100 over r stretchy right parenthesis end cell row P equals cell pi r plus 100 pi r to the power of negative 1 end exponent end cell end table

Differentiate P with respect to r

fraction numerator d P over denominator d r end fraction equals pi minus 100 pi r to the power of negative 2 end exponent

At the minimum point, the derivative will be zero (assuming the minimum is at a critical point)

table row cell pi minus 100 pi r to the power of negative 2 end exponent end cell equals 0 row cell pi minus fraction numerator 100 pi over denominator r squared end fraction end cell equals 0 row cell 1 minus 100 over r squared end cell equals 0 row 1 equals cell 100 over r squared end cell row cell r squared end cell equals 100 row r equals cell plus-or-minus 10 end cell end table

r is a length, so -10 can be ignored

r equals 10

Find the minimum value of the perimeter by substituting r equals 10 into the equation for P

P equals pi open parentheses 10 plus 100 over 10 close parentheses equals 20 pi

r equals 10 meters minimizes the perimeter

The minimum value of the perimeter will be 20π meters

(c) Justify that this is the minimum perimeter.

Answer:

To prove it is a minimum, show that the second derivative is positive (and the graph is therefore concave up) at this point

Start by finding the second derivative

fraction numerator d P over denominator d r end fraction equals pi minus 100 pi r to the power of negative 2 end exponent fraction numerator d squared P over denominator d r squared end fraction equals 200 pi r to the power of negative 3 end exponent

Substitute in r equals 10

right enclose fraction numerator d squared P over denominator d r squared end fraction end enclose subscript space r equals 10 end subscript equals 200 pi open parentheses 10 close parentheses to the power of negative 3 end exponent equals fraction numerator 200 pi over denominator 1000 end fraction equals pi over 5 greater than 0

Note that we don't need to check endpoints here: as r rightwards arrow 0 the perimeter function becomes unbounded, while as r rightwards arrow infinity the perimeter also goes to infinity

Second derivative is positive at r equals 10, therefore 20π  meters is the minimum value for the perimeter

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    Jamie Wood

    Author: Jamie Wood

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.