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Typical Steps for Solving Optimization Problems in Calculus I | MATH 115, Exams of Calculus

University of Michigan (UM) - Ann ArborCalculus

Material Type: Exam; Class: Calculus I; Subject: Mathematics; University: University of Michigan - Ann Arbor; Term: Winter 2007;

Typology: Exams

Pre 2010

Uploaded on 09/02/2009

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Typical Steps for Solving Optimization Problems, v. 3
March 25, 2007
Here is a list of steps that are often useful in solving optimization problems. A similar list appears in the
text on page 200.
1. Draw picture.
2. Label picture with variables. It is often easiest if you allocate a new variable name for each quantity
you want to label. We will later eliminate some variables.
3. Name the objective with a variable. (The objective is a technical term that refers to the quantity being
optimized or the associated function.)
4. Find the domains for all the labeled variables. (Finding the domain for xis a special case of finding a
constraint among xand other variables.)
5. Write objective in terms of other variables. At this point, you may find that you have not labeled
enough quantities. Label more, if necessary.
6. Which labeled quantities change and which are constant? Which are allowed to be present in the final
answer? Which labeled quantities may only change in a dependent way?
7. Find relationships among variables, from given constraints. If there are nvariables including the
objective, we generally need n−2 equations. It may help at this stage to identify tension among
intermediate goals.
8. Write objective in terms of just one variable. Usually, this means solving for all the auxiliary variables
in terms of just one variable.
9. Optimize the objective. Use the machinery from Chapter 4.3:
(a) Find the derivative of the objective function.
(b) Identify critical points—where the derivative is zero or undefined.
(c) Identify the endpoints of intervals in the domain of the objective function. Critical points and
endpoints are candidates for global extrema.
(d) Use the second derivative test or another such test to classify candidate points as local minima,
local maxima, or neither.
(e) If the domain is not closed (has some holes or excludes some endpoints) or goes off to −∞ and/or
+∞, then investigate the behavior of the function near the holes/endpoints/infinities. Usually
this means taking a limit, either algebraically or visually from a graph.
(f) The candidate point with the largest function value is the global maximum unless a bigger function
value can be gotten near an endpoint of the domain. (In the latter case, there is no global
maximum.)
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Typical Steps for Solving Optimization Problems, v. 3

March 25, 2007

Here is a list of steps that are often useful in solving optimization problems. A similar list appears in the text on page 200.

  1. Draw picture.
  2. Label picture with variables. It is often easiest if you allocate a new variable name for each quantity you want to label. We will later eliminate some variables.
  3. Name the objective with a variable. (The objective is a technical term that refers to the quantity being optimized or the associated function.)
  4. Find the domains for all the labeled variables. (Finding the domain for x is a special case of finding a constraint among x and other variables.)
  5. Write objective in terms of other variables. At this point, you may find that you have not labeled enough quantities. Label more, if necessary.
  6. Which labeled quantities change and which are constant? Which are allowed to be present in the final answer? Which labeled quantities may only change in a dependent way?
  7. Find relationships among variables, from given constraints. If there are n variables including the objective, we generally need n − 2 equations. It may help at this stage to identify tension among intermediate goals.
  8. Write objective in terms of just one variable. Usually, this means solving for all the auxiliary variables in terms of just one variable.
  9. Optimize the objective. Use the machinery from Chapter 4.3:

(a) Find the derivative of the objective function. (b) Identify critical points—where the derivative is zero or undefined. (c) Identify the endpoints of intervals in the domain of the objective function. Critical points and endpoints are candidates for global extrema. (d) Use the second derivative test or another such test to classify candidate points as local minima, local maxima, or neither. (e) If the domain is not closed (has some holes or excludes some endpoints) or goes off to −∞ and/or +∞, then investigate the behavior of the function near the holes/endpoints/infinities. Usually this means taking a limit, either algebraically or visually from a graph. (f) The candidate point with the largest function value is the global maximum unless a bigger function value can be gotten near an endpoint of the domain. (In the latter case, there is no global maximum.)

  1. Solve the problem asked. Which quantity is requested?
  2. Sanity check. Try a few other points to make sure that your point is optimum. Check units/dimensions.

We now repeat the list pointing out how these steps play out in 4.5 Example 1, a 40 in^3 can of minimal material.

  1. Draw picture. (See page 198.)
  2. Label picture with variables. It is often easiest if you allocate a new variable name for each quantity you want to label. We will later eliminate some variables. (See page 198. The variables are r and h.)
  3. Name the objective with a variable. (The objective is the amount of material, labeled by M .)
  4. Find the domains for all the labeled variables. (Here r ≥ 0 and h ≥ 0. The domain may be further restricted below.)
  5. Write objective in terms of other variables. At this point, you may find that you have not labeled enough quantities. Label more, if necessary. (We have M = 2πr^2 + 2πrh; see page 199.)
  6. Which labeled quantities change and which are constant? Which are allowed to be present in the final answer? Which labeled quantities may only change in a dependent way? (The quantites r, h, and M may change. We can control r and h (in a dependent way); the objective M depends on these, as usual.)
  7. Find relationships among variables, from given constraints. If there are n variables including the objective, we generally need n − 2 equations. It may help at this stage to identify tension among intermediate goals. (The constraint that the volume is 40 in^3 tells us that 40 = πr^2 h. Note that there are n = 3 variables, including the objective, namely, M, r, h. We have n − 2 = 1 relationship at this step. The objective is a second relationship and the optimization itself substitutes for a third relation among the three variables M, r, h. The tension here is that, to minimize material, we want to make both r and h small. But, if we make r smaller, we need to increase h in order to keep the volume at 40 in^3. See discussion on page 198. Sometimes one can conclude from the tension alone that there must be a unique local minimum. See also the discussion on line −3 of page 198 and the discussion on limits, below.) An alternative approach my have a fourth variable, V , for the volume of the can and a second equation, V = 40 in^3 .)
  8. Write objective in terms of just one variable. Usually, this means solving for all the auxiliary variables in terms of just one variable. (First we solve for h in terms of r, getting h = (^) πr^402. Then substitue this expression for h into the expression for M , getting M = 2πr^2 + (^80) r .)
  9. Optimize the objective. Use the machinery from Chapter 4.3:

(a) Find the derivative of the objective function. (We have dMdr = 4πr − (^80) r 2 .) (b) Identify critical points—where the derivative is zero or undefined. (The derivative is defined for all r 6 = 0. It is zero at r =

π

(c) Identify the endpoints of intervals in the domain of the objective function. (The domain is r > 0.) Critical points and endpoints are candidates for global extrema. (d) Use the second derivative test or another such test to classify candidate points as local minima, local maxima, or neither. (We have d

(^2) M dr^2 = 4π^ +^

160 r^3. Since^ r >^ 0, we conclude^

d^2 M dr^2 >^ 0.) Note that, to perform the second derivative test, we need to find the second derivative, but we don’t need to evaluate it. Avoiding the evaluation is probably good, because it might be a source of errors.