Geometric Optimization
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.7: “Applied Optimization Problems” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-7-applied-optimization-problems |
| Textbook used in class | Stewart, Calculus, Section 3.7: “Optimization Problems” |
Opening Scenario
Two geometry problems that appear in almost every Calculus 1 course:
- What shape of rectangle, among all rectangles with a fixed perimeter, has the largest area?
- What point on a given curve is closest to a given point not on the curve?
Both have clean answers. The rectangle of maximum area turns out to be a square. The closest point on a parabola to a given point requires minimizing a distance function.
These problems exercise the full setup-and-solve cycle, with the added geometric interpretation that makes the answers meaningful.
Key Concepts
1. Maximum Area Rectangle with Fixed Perimeter
Problem. Among all rectangles with perimeter $20$, find the one with the greatest area.
Setup. Let the sides be $l$ and $w$. Constraint: $2l + 2w = 20$, so $l + w = 10$ and $l = 10 - w$.
Objective: maximize $A = lw = (10-w)w = 10w - w^2$.
Domain: $0 < w < 10$.
Solve. $A'(w) = 10 - 2w = 0 \Rightarrow w = 5$. Then $l = 5$.
$A''(w) = -2 < 0$: local (and absolute) maximum.
$A(5) = 5 \times 5 = 25$.
Boxed answer: The rectangle with maximum area is a square with side $5$ and area $25$.
Geometric insight. Among all rectangles with a given perimeter, the square has the greatest area. This is a special case of the isoperimetric inequality: the shape that encloses the most area for a given perimeter is a circle.
2. Minimum Distance from a Point to a Curve
Problem. Find the point on the parabola $y = x^2$ that is closest to the point $(3, 0)$.
Setup. A point on the parabola has coordinates $(x, x^2)$. The square of the distance from $(x, x^2)$ to $(3, 0)$ is: $$D(x) = (x - 3)^2 + (x^2 - 0)^2 = (x-3)^2 + x^4.$$
We minimize $D(x)$ (minimizing the square of the distance minimizes the distance, since the square root is an increasing function).
Domain: all real $x$.
Solve. $D'(x) = 2(x-3) + 4x^3 = 4x^3 + 2x - 6$.
Factor: try $x = 1$: $4 + 2 - 6 = 0$. So $(x-1)$ is a factor.
$D'(x) = (x-1)(4x^2 + 4x + 6)$.
The quadratic $4x^2 + 4x + 6$ has discriminant $16 - 96 < 0$: no real roots. So the only real critical number is $x = 1$.
Classify. $D'(x) < 0$ for $x < 1$ and $D'(x) > 0$ for $x > 1$ (check: at $x = 0$, $D'(0) = -6 < 0$; at $x = 2$, $D'(2) = 32 + 4 - 6 = 30 > 0$). Sign goes from $-$ to $+$: local and absolute minimum.
The minimum distance point is $(1, 1)$.
Distance $= \sqrt{D(1)} = \sqrt{(1-3)^2 + 1} = \sqrt{4 + 1} = \sqrt{5}$.
Boxed answer: The closest point on $y = x^2$ to $(3, 0)$ is $(1, 1)$. The minimum distance is $\sqrt{5}$.
Why minimize $D$ instead of the distance? The actual distance is $\sqrt{D}$. Since $\sqrt{\cdot}$ is increasing, the minimum of $\sqrt{D}$ occurs at the same $x$ as the minimum of $D$. Working with $D$ avoids differentiating a square root.
3. Rectangle Inscribed in a Semicircle
Problem. Find the rectangle of largest area that can be inscribed in a semicircle of radius $r$.
Setup. Place the semicircle with diameter on the $x$-axis and center at the origin. The semicircle is the curve $y = \sqrt{r^2 - x^2}$ for $x \in [-r, r]$. A rectangle inscribed in the semicircle has its base on the diameter and its upper two corners on the semicircle.
By symmetry, let the upper-right corner be at $(x, \sqrt{r^2 - x^2})$ with $x > 0$. Then:
- Width of rectangle: $2x$.
- Height of rectangle: $\sqrt{r^2 - x^2}$.
- Area: $A(x) = 2x\sqrt{r^2 - x^2}$, for $0 < x < r$.
Solve. It is easier to maximize $A^2 = 4x^2(r^2 - x^2)$. Let $u = x^2$:
$(A^2)' = 4[2x(r^2 - x^2) + x^2(-2x)] = 4[2xr^2 - 2x^3 - 2x^3] = 4x[2r^2 - 4x^2]$.
Setting $2r^2 - 4x^2 = 0$: $x^2 = r^2/2$, so $x = r/\sqrt{2}$.
$A\!\left(\dfrac{r}{\sqrt{2}}\right) = 2 \cdot \dfrac{r}{\sqrt{2}} \cdot \sqrt{r^2 - r^2/2} = \dfrac{2r}{\sqrt{2}} \cdot \dfrac{r}{\sqrt{2}} = \dfrac{2r^2}{2} = r^2$.
Boxed answer: The largest inscribed rectangle has area $r^2$, with width $r\sqrt{2}$ and height $r/\sqrt{2}$.
Geometric note. The width equals the height times $\sqrt{2}$: the rectangle has side ratio $\sqrt{2}:1$, not a square. The square is optimal for a fixed perimeter; the half-square ($\sqrt{2}:1$ rectangle) is optimal for the inscribed problem.
4. Predict-Then-Check Practice
Before computing: For the rectangle inscribed in a semicircle of radius $2$, predict whether the optimal width is more or less than the radius.
Prediction. The rectangle must fit inside the semicircle, so the width must be less than $2r = 4$. A very wide, flat rectangle wastes height; a very narrow, tall rectangle wastes width. The optimum should be somewhere in between. Width $= r\sqrt{2} = 2\sqrt{2} \approx 2.83 < 4$: yes, between $0$ and $4$.
Check. From the formula: width $= r\sqrt{2} = 2\sqrt{2} \approx 2.83$. Prediction confirmed as reasonable.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Minimizing the square of the distance but reporting $D(c)$ as the distance | Saying the min distance is $5$ when $D(1) = 5$ | The min distance is $\sqrt{D(c)} = \sqrt{5}$; check whether you are minimizing distance or distance-squared |
| Forgetting to check that the critical number gives the correct extremum type | Differentiating and stopping at the critical number | Test the sign of $A''$ or use the FDT to confirm max or min |
| Using the wrong formula for area of the inscribed rectangle | Writing $A = x \cdot \sqrt{r^2 - x^2}$ instead of $2x \cdot \sqrt{r^2 - x^2}$ | The base of the symmetric rectangle spans from $-x$ to $x$, so width is $2x$ |
Common Misconceptions
minimizing the distance to a point is equivalent to minimizing the square of the distance and then taking the square root of the objective function’s value at the critical number.
This is the concept-image-conflicts-definition error about what the optimized quantity represents. Setting up $D(x) = (x-3)^2 + x^4$ and finding its minimum value $D(1) = 5$ gives the minimum squared distance. The actual minimum distance is $\sqrt{D(1)} = \sqrt{5}$, not $5$. The shortcut of minimizing $D$ is valid because the square root is an increasing function, so the minimizer is the same, but the minimum value must be square-rooted before reporting a distance.
the rectangle of maximum area inscribed in any symmetric shape is always a square.
This is the concept-image-conflicts-definition error about which geometric optimum applies. A square maximizes area among rectangles with fixed perimeter, but the inscribed problem imposes a different constraint. For a semicircle of radius $r$, the optimal inscribed rectangle has side ratio $\sqrt{2}:1$, not $1:1$. The specific shape of the bounding curve determines which proportions are optimal; the result must be derived for each setup rather than assumed from the fixed-perimeter case.
Leveled Practice
Level 1 -- Direct Application
Problem 1. Among all rectangles with area $64$, find the one with the minimum perimeter.
Show answer
Let sides be $l$ and $w$. Constraint: $lw = 64$, so $l = 64/w$.
Objective: minimize $P = 2l + 2w = 128/w + 2w$.
$P'(w) = -128/w^2 + 2 = 0 \Rightarrow w^2 = 64 \Rightarrow w = 8$. Then $l = 8$.
$P''(w) = 256/w^3 > 0$: minimum.
Boxed answer: The square with side $8$ has minimum perimeter $32$.
Level 2 -- Multiple Steps
Problem 2. Find the point on the line $y = x + 2$ closest to the origin.
Show answer
A point on the line is $(x, x+2)$. Square of distance from origin: $D(x) = x^2 + (x+2)^2 = 2x^2 + 4x + 4$.
$D'(x) = 4x + 4 = 0 \Rightarrow x = -1$.
$D''(x) = 4 > 0$: minimum.
Closest point: $(-1, 1)$. Distance: $\sqrt{(-1)^2 + 1^2} = \sqrt{2}$.
Boxed answer: The closest point on $y = x + 2$ to the origin is $(-1, 1)$. Minimum distance is $\sqrt{2}$.
Mastery Checklist
Mental Model
Geometric optimization problems always reduce to the same two-step pattern:
Step 1 (geometry): Translate the geometry into a function. Use coordinates, formulas for area, volume, perimeter, and distance. Eliminate extra variables with the constraint. This step is pure algebra and trigonometry.
Step 2 (calculus): Find the critical number of the resulting function and classify it. This step is the same calculation you have done throughout Chapter 3.
The calculus is not the hard part. Building the correct objective function is.
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