Optimization, linear approximation and differentials

The most “applied” week of Calculus I. In an optimization problem you turn a word situation into a function and find the best value with the extrema knowledge of week 8. Linear approximation is the use of the derivative’s idea “the tangent line is close to the function” for estimating numbers without a calculator. Differentials are the application of the same idea to error computation. Things to keep in mind:

1. In optimization the hard part is not the derivative, it is setting up the function. Objective function + constraint.
2. Use the constraint to reduce to one variable. You cannot differentiate a function of two variables.
3. Show that the point you found is a max or a min. The endpoints or the $f''$ test.
4. Linear approximation = the tangent line. $L(x)=f(a)+f'(a)(x-a)$.
5. $dy$ is the change along the tangent, $\Delta y$ is the actual change.

Optimization problems

The six-step method

1. Draw a figure, name the variables, write the units.
2. Objective function: what are you maximizing/minimizing? (area, volume, cost, distance)
3. Constraint: the information linking the variables (perimeter 600, volume 4000...)
4. Solve the constraint for one variable, substitute into the objective -> a single-variable f(x).
5. Write the interval that makes physical sense (such as length > 0).
6. Solve f'(x) = 0; verify that it is a max/min with the endpoints or with f''.

Example 1 (pen). A rectangular area will be enclosed with $600$ m of fence and divided into two equal compartments by a fence parallel to one side. Maximize the total area.

Let the dividing fence be parallel to the sides of length $x$. Then there are three fences of length $x$ (two outer sides and the divider), and two fences of length $y$:

$3x+2y=600 \Rightarrow y=\frac{600-3x}{2}$

Which side you choose to be the triple one does not change the result; only the labels swap. Drawing the figure and counting the fences one by one prevents errors at this step.

$A(x)=x\cdot\frac{600-3x}{2}=300x-\frac32x^2,\qquad 0<x<200$

$A'(x)=300-3x=0 \Rightarrow x=100$. $A''=-3<0$, max. $y=\dfrac{600-300}{2}=150$. The largest area is $100\cdot150=15000$ m².

Example 2 (material). The least material for a box with a square base, an open top and a volume of $4000$ cm³.

base side x, height h
objective:   S = x^2 + 4xh                 (base + 4 side faces, no top)
constraint:  x^2 h = 4000  ->  h = 4000 / x^2
S(x) = x^2 + 16000 / x,   x > 0
S'(x) = 2x - 16000 / x^2 = 0  ->  x^3 = 8000  ->  x = 20
S''(x) = 2 + 32000 / x^3 > 0  ->  minimum
h = 4000 / 400 = 10,   S = 400 + 800 = 1200 cm^2

On an open interval ($x>0$) there are no endpoints. Since there is a single critical point and $S''>0$, this local min is also the absolute min. Write this sentence on the exam.

PITFALL · not checking whether it is a max or a min. $f'=0$ gives a candidate. Before writing the answer, either compare with the endpoint values or show the sign of $f''$. For example in the wire problem (Question 12) the critical point gives the minimum; the maximum is at an endpoint, when all the wire goes to the circle.

TIP · use the square for distances. Instead of $d=\sqrt{(x-a)^2+(y-b)^2}$, minimize $d^2$. The square root is an increasing function, and the smallest $d$ and the smallest $d^2$ are at the same point. The derivative is much shorter.

Linear approximation

If $f$ is differentiable at $a$, then for $x$ near $a$ the graph is very close to the tangent line:

$f(x)\approx L(x)=f(a)+f'(a)(x-a)$

$L$ is called the linearization of $f$ at $a$. Choose $a$ as a point close to the target number for which you can compute $f(a)$ and $f'(a)$ in your head.

Example 3. Estimating $\sqrt{9.2}$. $f(x)=\sqrt x$, $a=9$: $f(9)=3$, $f'(9)=\dfrac16$.

$L(9.2)=3+\frac16(0.2)=3.0333\ldots\qquad(\text{true: }3.03315\ldots)$

Commonly used approximations ($x$ near zero, $a=0$):

Example 4. $(1.02)^5\approx1+5(0.02)=1.1$. The true value is $1.10408$. The error is of the order of the square of $x=0.02$.

Over or under? If $f''<0$ (concave down), the graph stays below the tangent and $L$ overestimates (as for $\sqrt x$). If $f''>0$, it underestimates (as for $e^x$).

Differentials

For $y=f(x)$, $dx$ is an independent variable (a small increment), and

$dy=f'(x)\,dx$

$dy$ is the rise along the tangent line. The actual change is $\Delta y=f(x+dx)-f(x)$. When $dx$ is small, $dy\approx\Delta y$.

Example 5. $f(x)=x^2$, $x=3$, $dx=0.1$: $dy=2x\,dx=0.6$, $\Delta y=3.1^2-9=0.61$. The difference is exactly $dx^2=0.01$.

Error estimation

If you measure one quantity and compute another from it, the measurement error propagates to the computed value. The differential estimates it.

Example 6. The edge of a cube is $10$ cm, with a measurement error of $\pm0.1$ cm.

V = x^3        dV = 3x^2 dx = 3 · 100 · 0.1 = 30 cm^3     (absolute error)
relative error  = dV / V = 30 / 1000 = 0.03                (3 percent)
shortcut:        dV / V = 3 · dx / x = 3 · 0.01 = 0.03

General rule: if $Q=kx^n$, then $\dfrac{dQ}{Q}=n\dfrac{dx}{x}$. The larger the power, the larger the relative error.

PITFALL · confusing $dy$ and $\Delta y$. If the question says “use differentials to estimate”, compute $dy=f'(x)dx$. If it says “find the actual change”, compute $f(x+\Delta x)-f(x)$. They may ask for both together and also ask for the difference.

Common mistakes