Optimization

Contents

General

  1. Read questions carefully
  2. Draw one or more diagrams. Label given quantities.
  3. Define suitable variables. Use ones that belong in a formula.
  4. Reduce to a single expression to optimize
  5. Write down domain for the expression
  6. Find endpoint extrema (if any) and critical points.
  7. Find the desired max or min
  8. Answer the question (does question want xx, yy or something else)
  9. Check if answer makes sense

Angles

Calculus of trig functions are only done in radians. If an optimization problem involves angles in degrees, then first convert them to radians before doing calculus.

Ratio of functions

If, say from applying a quotient rule, you have

f(x)g(x)=0\frac{f(x)}{g(x)} = 0

then you can simply set f(x)f(x) to 00, as long as you are sure that ff and gg do not share a common root.

One-to-one functions

If ff is a one-to-one (ie invertible), continuous function over the range of gg, then

 d dx(f∘g)(x)=0\frac{\d}{\d x} (f \circ g)(x) = 0

has the same critical points as

 d dxg(x)=0\frac{\d}{\d x} g(x) = 0

For example, instead of directly minimizing (or maximizing) distance involving a square root, you can simply minimize (or maximize) the expression under the square root.

Example

Example: A person is stuck in the water 30 m30 \text{ m} away from the beach. The lifeguard on the beach is 20 m20 \text{ m} from the water, but is 60 m60 \text{ m} further down the beach than the person. The lifeguard can travel 4.5 m s−14.5 \text{ m s}^{-1} on the beach, and 1.5 m s−11.5 \text{ m s}^{-1} in the water. Find the fastest time that the lifeguard could take to get to the person.

The lifeguard should travel in a straight line to the water, then in a straight line to the person. Let xx be the distance in meters the lifeguard traveled along the beach when they get to the water, such that 0≤x≤600 \leq x \leq 60. The lifeguard would travel 202+x2 m\sqrt{20^2 + x^2}\text{ m} on the beach and 302+(60−x)2 m\sqrt{30^2 + (60-x)^2}\text{ m} in water. Let tt be the time taken in seconds.

time=distancespeedt(x)=202+x24.5+302+(60−x)21.5\begin{align*} \text{time} &= \frac{\text{distance}}{\text{speed}} \\ t(x) &= \frac{\sqrt{20^2 + x^2}}{4.5} + \frac{\sqrt{30^2 + (60-x)^2}}{1.5}\end{align*}

The times for the two straight-line paths were added up. We continue using graphing calculator.

 dt dx∣x0=0x0=50.229…t(50.229)≈33.0\begin{align*} \frac{\d t}{\d x} \Big\vert_{x_0} &= 0 \\ x_0 &= 50.229 \dots \\ t(50.229) &\approx 33.0 \end{align*}

Since 0≤x≤600 \leq x \leq 60, we also check that t(60)≈34.1t(60) \approx 34.1 and t(0)=49.2t(0) = 49.2 which are both greater than our critical value. So t≈33.0 st \approx 33.0 \text{ s} is the minimum time ■\qed

Interestingly, the beach to water speed ratio is 3.03.0 and the x0202+x02\frac{x_0}{\sqrt{20^2 + x_0^2}} to 60−x0302+(60−x0)2\frac{60 - x_0}{\sqrt{30^2 + (60-x_0)^2}} ratio is also 3.03.0. In physics, this is called Snell’s law.

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