7. Optimization and Approximation

Modeling Optimization Problems

Optimization asks for the largest or smallest possible value of an objective quantity while respecting one or more constraints. Common objectives include area, volume, revenue, profit, distance, cost, and height.

A useful modeling sequence is:

1. Identify the objective quantity.

2. Define variables for the unknown quantities.

3. Write the constraint equation.

4. Use the constraint to express the objective with one independent variable when possible.

5. Determine the meaningful domain.

6. Find critical points by solving $f'(x)=0$ and by checking where $f'(x)$ does not exist.

7. Compare all candidate values and relevant endpoints.

8. Interpret the result with units and context.

For a rectangle with perimeter $40$ meters, let the side lengths be $x$ and $y$. The constraint is $2x+2y=40$, so $y=20-x$. The area becomes the one-variable function $A(x)=x(20-x)=20x-x^2$, with domain $0\le x\le20$. Since $A'(x)=20-2x$, the interior candidate is $x=10$, giving $y=10$. Because $A''(x)=-2<0$, the rectangle with greatest area is a square measuring $10$ meters by $10$ meters.

A derivative equal to zero identifies a possible optimum, not automatically an actual maximum or minimum. The domain and comparison of candidates are essential.

Takeaway: Translate the context into a constrained one-variable function before differentiating.

Absolute Extrema on Closed Intervals

The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval $[a,b]$ has an absolute maximum and an absolute minimum. These values occur either at an endpoint or at an interior critical point.

To find absolute extrema:

1. Evaluate $f(a)$ and $f(b)$.

2. Find every critical point in $(a,b)$, where $f'(x)=0$ or $f'(x)$ is undefined.

3. Evaluate the function at every critical point in the interval.

4. Compare all resulting values.

5. Identify the largest as the absolute maximum and the smallest as the absolute minimum.

For $f(x)=x^3-3x$ on $[-2,2]$, the derivative is $f'(x)=3x^2-3=3(x-1)(x+1)$, so the critical points are $x=-1$ and $x=1$. The relevant values are $f(-2)=-2$, $f(-1)=2$, $f(1)=-2$, and $f(2)=2$. Therefore, the absolute maximum is $2$, occurring at $x=-1$ and $x=2$, while the absolute minimum is $-2$, occurring at $x=-2$ and $x=1$.

Takeaway: Never omit endpoints in an absolute-extrema problem.

Applications in Geometry, Economics, and Physics

Optimization models appear in geometry, economics, and physical applications. In each case, the derivative supplies local information that helps locate a best value.

For a cylindrical can with fixed volume $V$, the constraint is $\pi r^2h=V$, so $h=\frac{V}{\pi r^2}$. Substituting into the surface-area formula $S=2\pi r^2+2\pi rh$ produces a function of $r$ alone. Its critical points can identify the radius that uses the least material.

In economics, if $q$ is the quantity sold and $p(q)$ is the price per unit, revenue is $R(q)=q\,p(q)$, and profit is $P(q)=R(q)-C(q)$. For $p(q)=100-2q$ and $C(q)=20q+100$, profit is

Since $P'(q)=80-4q$, the candidate is $q=20$. The second derivative $P''(q)=-4<0$ confirms maximum profit at that quantity. The corresponding price is $p(20)=60$, and the maximum profit is $P(20)=700$. The domain must still ensure that quantity and price are meaningful.

For a projectile with height h(t)=-16t^2+64t+5\, vertical velocity is $h'(t)=-32t+64$. Setting velocity to zero gives $t=2$, and the maximum height is $h(2)=69$ feet.

Takeaway: The same workflow applies across fields, but the mathematical domain must remain consistent with the real situation.

Linear Approximation

A linearization replaces a differentiable function near $x=a$ with its tangent line:

For inputs close to $a$, use $f(x)\approx L(x)$. This is effective when $f(a)$ and $f'(a)$ are easy to calculate but the exact value of $f(x)$ is inconvenient.

To approximate $\sqrt{4.1}$, choose $f(x)=\sqrt{x}$ and base point $a=4$. Then $f(4)=2$ and $f'(4)=\frac14$, so

Therefore,

The estimate is close because $4.1$ is near the base point $4$. The approximation generally becomes less reliable as the input moves farther from the point where the tangent line was constructed.

Takeaway: Linear approximation is a local method: select a convenient nearby point, compute the tangent line, and use it only within a suitable neighborhood.

Differentials and Error Estimates

A differential estimates the change in a function caused by a small change in its input. If $y=f(x)$, then

The exact change is

whereas $dy$ gives an approximation when $dx=\Delta x$ is small:

For the area of a circle, $A=\pi r^2$, so $dA=2\pi r\,dr$. If $r=10$ centimeters and $dr=0.02$ centimeters, then

Thus, the area increases by approximately $1.257$ square centimeters. The same calculation can estimate how a small measurement error in radius affects the calculated area.

Takeaway: Differentials convert a small input change into an approximate output change through the derivative.

Newton’s Method

Newton’s method repeatedly applies tangent-line approximations to solve an equation. To solve $f(x)=0$, begin with an initial guess $x_0$ and use

Geometrically, the tangent line at $(x_n,f(x_n))$ meets the horizontal axis at the next approximation.

To approximate $\sqrt{10}$, solve $x^2-10=0$. With $f(x)=x^2-10$ and $f'(x)=2x$, the iteration becomes

Starting with $x_0=3$ gives

Hence, $\sqrt{10}\approx3.16227766$.

The method is not guaranteed to converge. Difficulties can arise if $f'(x_n)=0$, the initial guess is far from the desired root, an iteration enters an unsuitable domain, or the function has multiple roots or complicated behavior. A practical stopping rule is

combined with checking the residual $|f(x_n)|$ to ensure that the approximation nearly satisfies the original equation.

Takeaway: Newton’s method is powerful but depends on a suitable starting point, a defined derivative, and a verification of the result.

Connecting Optimization and Approximation

These methods use the derivative in related ways. Optimization uses derivative information to locate candidate extrema. Absolute-extrema analysis adds endpoints to the critical-point test. Linear approximation replaces a function near one point with its tangent line. Differentials express the resulting approximate change, and Newton’s method repeats tangent-line approximation to find a root.

The shared pattern is local information supporting a larger decision:

• To optimize, compare derivative-based candidates with endpoints.

• To estimate a value, use a tangent line near a convenient base point.

• To estimate a change, use $dy=f'(x)\,dx$.

• To solve an equation, update an approximation with the Newton iteration.

Each method has conditions and limitations. Check continuity, differentiability, domains, endpoint behavior, the distance from a linearization point, and the quality of an iterative approximation.

Final takeaway: Derivatives do more than describe instantaneous change; they provide a systematic way to model best values, estimate nearby quantities, and approximate solutions.