Math · Calculus I · Concept
Optimization problems in calculus
To solve an optimization problem, write the quantity to maximize or minimize as a function of one variable, using the constraint to remove any other variable. Find the critical points where the derivative is zero, then compare the function’s values there and at the endpoints of the allowed interval.
Turn the words into one function
Name the quantity to optimize and every other quantity with a variable. A constraint, such as a fixed length of fence, links the variables: solve it for one variable and substitute, so the quantity depends on a single variable.
Find the critical points
A maximum or minimum inside an interval can occur only where the derivative is zero or undefined. Differentiate, set the derivative equal to zero and solve.
Check the endpoints too
On a closed interval, the largest or smallest value can sit at an endpoint. Compare the function at every critical point and at both endpoints: this is the closed interval method. For the fence, A(0) = A(120) = 0, so x = 60 gives the maximum.
Or use the second derivative
At a critical point, a negative second derivative means the graph is concave down there, so the point is a local maximum; a positive one means a local minimum. When it is the only critical point in the interval, that local answer is also the overall one.
Answer the question that was asked
Report the quantity the problem asks for, with units: the dimensions, the largest area or the least cost. Check that the answer makes sense, with lengths positive and the constraint satisfied.
Common mistakes
- Differentiating before using the constraint, so the function still has two variables.
- Forgetting the endpoints, where the largest or smallest value can occur.
- Stopping at x = 60 when the question asks for the area or both dimensions.
- Keeping a critical point outside the allowed interval, such as a negative length.
Key terms
- Optimization
- Finding the largest or smallest value of a quantity, often under a constraint. In calculus the candidates are the critical points inside the allowed interval and its endpoints.
- Critical point
- A point in the function’s domain where the derivative is zero or doesn’t exist. Local maxima and minima can happen only at critical points or endpoints, so these are where to look.
- Extremum
- A maximum or minimum value. A local extremum is highest or lowest among nearby values; an absolute extremum is highest or lowest over the whole domain, and it can sit at an endpoint where f′ isn’t zero.
- Second derivative test
- At a critical point c with f′(c) = 0, f″(c) < 0 gives a local maximum and f″(c) > 0 a local minimum. If f″(c) = 0 the test gives no answer, and the first derivative test decides.
- Derivative
- The instantaneous rate of change of a function: the limit of the average rate of change as the step shrinks to zero, when that limit exists. On a graph it is the slope of the tangent line.
Work through an example
A farmer has 240 m of fencing to enclose a rectangular field along a straight river, with no fence on the river side. What dimensions give the largest area?
Maximize the area of a fenced field →Sources and scope
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Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
Graph the area in Graph Check each step in Math See the zero slope in Slopes, sums & signed area Open worked example on a board Optimization candidates in Math ReferenceYour existing work stays on this device. Examples open as editable copies.