Calculus Reference

Optimization

Optimization problems use derivatives to find the largest or smallest value allowed by a constraint.

Solved Examples

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Optimization Method

  1. Define the quantity to maximize or minimize.
  2. Write the constraint equation.
  3. Rewrite the target as one variable.
  4. Find critical numbers and check endpoints.
  5. State the dimensions and the requested extreme value.

Maximize an Area with a Fixed Perimeter

Rectangle with Perimeter $20$

Find the dimensions of the rectangle with perimeter $20$ and greatest area.

Solution Steps
  1. Use the perimeter constraint to write $y$ in terms of $x$.
  2. Write the area as a one-variable function.
  3. Set the derivative equal to zero and solve for $x$.
  4. Use the constraint to find $y$.
  5. Evaluate the area at the critical dimensions.
  6. State the maximum area and dimensions.
$$2x+2y=20\quad\Rightarrow\quad y=10-x$$$$A(x)=x(10-x)=10x-x^2$$$$A'(x)=10-2x=0\quad\Rightarrow\quad x=5$$$$y=10-5=5$$$$A(5)=25$$$$\text{maximum area: }25\text{ square units, from a }5\times5\text{ square}$$

Use a Context Constraint

Fence Along a Wall

A wall forms one side of a rectangular garden. There are $24$ meters of fence for the other three sides. Find the dimensions with greatest area.

Solution Steps
  1. Let $x$ be each side perpendicular to the wall and $y$ the side parallel to it.
  2. Use the three-sided fencing constraint.
  3. Write area as a one-variable function.
  4. Find the critical number and recover the second dimension.
  5. State the maximum area in context.
$$2x+y=24\quad\Rightarrow\quad y=24-2x$$$$A(x)=x(24-2x)=24x-2x^2$$$$A'(x)=24-4x=0\quad\Rightarrow\quad x=6$$$$y=24-2(6)=12$$$$\boxed{\text{maximum area }72\text{ m}^2\text{ from a }6\text{ m by }12\text{ m garden}}$$

Check a Closed Interval

Absolute Maximum and Minimum

Find the extreme values of $f(x)=x^2-4x+3$ on $[0,4]$.

Solution Steps
  1. Find the critical number inside the interval.
  2. Evaluate the function at the critical number and both endpoints.
  3. State the absolute minimum.
  4. State the absolute maximum.
$$f'(x)=2x-4=0\quad\Rightarrow\quad x=2$$$$f(0)=3,\qquad f(2)=-1,\qquad f(4)=3$$$$\text{absolute minimum: }-1\text{ at }x=2$$$$\text{absolute maximum: }3\text{ at }x=0\text{ and }x=4$$

Minimize a Cost Model

Balance Two Costs

For $x>0$, a cost model is $C(x)=x+\frac{16}{x}$. Find the minimum cost.

Solution Steps
  1. Differentiate the one-variable cost model.
  2. Set the derivative equal to zero.
  3. Use the positive-domain condition.
  4. Evaluate the cost at the critical input.
  5. State the minimum value and where it occurs.
$$C'(x)=1-\frac{16}{x^2}$$$$1-\frac{16}{x^2}=0\quad\Rightarrow\quad x^2=16$$$$x=4\quad\text{because }x>0$$$$C(4)=4+\frac{16}{4}=8$$$$\boxed{\text{minimum cost }8\text{ at }x=4}$$