Optimization Problems in Calculus: How to Solve Them

You have 500 feet of fence and a building to use as one side. Where do the posts go to enclose the most ground? Guess and you will be close. Differentiate and you will be exact. One common application of calculus is calculating the minimum or maximum value of a function¹. Companies minimize production costs or maximize revenue, and in manufacturing it is often desirable to minimize the material used to package a product with a certain volume¹.

What is an optimization problem?

In an optimization problem we look for the largest value or the smallest value that a function can take². The version in this chapter adds a twist: we look for the largest or smallest value of a function subject to some kind of constraint². A particular quantity is the one we want to maximize or minimize, and there is also some auxiliary condition that must be satisfied¹.

The constraint is a condition, usually an equation, that must absolutely, positively be true no matter what our solution is². In almost every problem one quantity is clearly indicated as having a fixed value, and that is the constraint². In the wider world the subject is large: mathematical optimization is the selection of a best element, with regard to some criteria, from some set of available alternatives³. Calculus handles the smooth, continuous case. In the general approach, an optimization problem maximizes or minimizes a real function by choosing input values from within an allowed set³.

Optimization is the art of letting a constraint do the choosing, so the derivative only has one thing left to decide.

How to solve an optimization problem: the method

The first step in all of these problems, per the Lamar University notes, is to read the problem very carefully². Then the next step is to identify the quantity to be optimized and the constraint². The remaining steps:

• Draw it. If applicable, draw a figure and label all variables¹. Sketching the situation will often help to arrive at the equations².
• Name the target. Determine which quantity is to be maximized or minimized, and for what range of values of the other variables¹.
• Write the target as a formula. Write a formula for the quantity to be maximized or minimized in terms of the variables¹.
• Use the constraint to get one variable. There are 2 functions in every problem, the one to optimize and the constraint². We know how to find the largest or smallest value of a function provided it has a single variable, so solve the constraint for one variable and substitute².
• Find the domain. Both sides of a physical shape must be positive, for example¹.
• Locate the optimum. This step typically involves looking for critical points and evaluating a function at endpoints¹. The maximum value of a continuous function on a closed interval will occur at critical points and/or end points².
• Answer the question asked. Do not forget to actually read the problem and give the answer that was asked for².

Worked example 1: the fenced field

The problem is the standard one: enclose a rectangular field with a fence, with 500 feet of fencing material and a building on one side that needs no fence, and find the dimensions that enclose the largest area². The source gives the answer, 250 by 125². The arithmetic below is ours and gets there:

Worked example (our arithmetic): let x be the width of each of the 2 sides perpendicular to the building and y the side parallel to it¹.

The diagram above is this function: the area climbs to its peak at x = 125 and falls back to 0, and the slope is zero exactly at the top². The best field is half as wide as it is long.

How to be sure it is the maximum

Finding a critical point is not the same as proving it is the best one. The first derivative test identifies points that might be extrema, but it does not by itself say whether a point is a minimum, a maximum or neither³. A problem on a closed interval gets its answer by comparing the values at the critical points and at the endpoints². Fermat’s work gives the underlying idea: optima of unconstrained problems are found at stationary points, where the first derivative is zero³. Fermat and Lagrange found calculus-based formulae for identifying optima³.

There are other checks. Not every problem has a closed interval with finite endpoints, and many of the functions are not continuous once reduced to one variable². The second derivative can help: it is possible to use it to exclude values, though by itself that does not identify the optimum². Problems with more than one critical point are often difficult, because it is hard to know which critical point gives the optimal value². In the examples below the vast majority have a single critical point². A function can have only a single absolute maximum value, but that value can occur at more than one place in the domain, and some graphs have minimums but no maximums, as the Lamar University notes on absolute extrema show⁴. Whatever method you use, you always need some method for making sure the answer is the true absolute extreme². In the fence example the endpoints made no sense, so only the critical point was left².

Worked example 2: the open-top box

Start with a flat sheet of cardboard 14 inches by 10 inches. Cut a square from each corner, fold up the flaps, and you have an open-top box. The setup is the textbook cardboard problem, with the strategy applied to maximize the volume of the open-top box¹. Worked example (our arithmetic): the side of the cut-out square is x inches¹.

Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial¹. Notice the step where one root was thrown out because it fell outside the domain. Do not get into the habit of just excluding any negative critical point, though, since there are problems where negative critical points are perfectly valid solutions².

Worked example 3: revenue

Business is the other great source of these problems. Companies are interested in maximizing revenue¹, and a car rental company has to balance the price of a rental against the number of cars people will rent at that price¹. The revenue per day is the number of cars rented per day times the price charged per car per day¹. Our numbers are invented for this example, and the method is the textbook’s¹. Worked example (our arithmetic): the number of cars rented is N = 120 − 2p when the price is p dollars per day, and the owners will charge between 20 and 50 dollars.

Both endpoints give less than the critical point, so the interior point wins. If the interval had been 40 to 50 dollars, the best price would have been 40, at an endpoint, and the derivative would never have been zero in range¹.

Worked example 4: the can

A manufacturer needs a cylindrical can that will hold 1.5 liters of liquid and wants to minimize the material used². Because the radius and height are lengths, convert the volume to cubic centimeters: 1.5 liters is 1,500 cubic centimeters². Here the volume is the constraint and the surface area is the quantity to minimize². The source finds that a radius of 6.2035 cm and a height of 12.4070 cm use the least material². Worked example (our arithmetic), reaching the same numbers:

The best can is as tall as it is wide. A can that holds the same volume but has a radius of 5 cm would need a height of 19.1 cm and 3000 / 5 + 2π × 25 = 757.1 square cm of material, more than 725.4 (our arithmetic)².

Where students go wrong

This section is generally one of the more difficult for students², and the reasons are not about calculus. A subtle change of wording can completely change the problem². There is also the problem of identifying the quantity to optimize and the quantity that is the constraint, and of writing down equations for each².

• Locking into one template. Many problems can be done in both directions, with the roles of the 2 quantities swapped². It is important not to get so locked into one way of doing these problems that you cannot do them the opposite way². This is one of the more common mistakes students make².
• Trusting one verification method. Two different methods of verifying the optimal value were used in the textbook’s examples, so do not get locked into one².
• Forgetting units. The can example needed a liter-to-cubic-centimeter conversion before any calculus².
• Answering the wrong question. Read the problem again before writing the final line².
• Skipping the domain. The domain decides which critical points count¹.

Practice problems in the style of section 4.7

Try these before checking the answers (our arithmetic, same method as above)¹:

Beyond the classroom

Optimization reaches well past calculus class. In discrete optimization, an object such as an integer, permutation or graph must be found from a countable set³. Calculus-style methods, with their derivatives, belong to the continuous case³. The derivative-is-zero idea you used here is the same one at work in larger problems: Newton and Gauss proposed iterative methods for moving towards an optimum³.

Where to go next

Optimization needs the derivative, so read what a derivative is if the idea of a zero slope at the peak is new. Messy functions often need the chain rule, and the derivative rules cheat sheet lists the rest. The core ideas section has the other building blocks. For more worked problems, see Lamar University’s optimization notes and OpenStax section 4.7.