Math Class Didn’t Show Its Work, companion reader · Chapter 30

Where Is the Best Point?

The best point is never best until the question says best for what, and over which inputs.

These four demonstrations follow the chapter's rectangle with perimeter 20 from its first sketch to its final check, then move the boundary of a second model to see when the turning point stops counting. Every number is small enough to check with a pencil.

Every example in these readers is a constructed teaching example taken from or modeled on the book's own worked examples. The numbers are declared inputs chosen to make the mathematics visible. They are not measurements of any real person, product or study.

Demonstration 1 of 4

One constraint, one variable

If the perimeter must be 20, how much choice is really left?

Once x is chosen, the perimeter decides y for you. The left picture is the rectangle; the right picture places its area on the curve A(x) = x(10 - x).

\[y = 10 - x\]

\[A(x) = xy\]

x and y are the two side lengths in units. The perimeter 2x + 2y = 20 becomes y = 10 - x. A(x) is the area in square units.

Predict first. Before you pick x = 2, predict the other side and the area. Is it more or less than 25?

Choose an example

Figure: One constraint, one variable. Left: a 5 by 5 shape drawn to scale. Right: the area curve A(x) = x(10 - x) from 0 to 10 with the point x = 5, area 25, marked.
Side x (units): 5
Constructed example: the chapter's 1 by 9, 2 by 8 and 5 by 5 rectangles, plus its x = 10 boundary.

Calculated values

Side x
5 units
Other side y = 10 - x
5 units
Perimeter
2 x 5 + 2 x 5 = 20 units
Area A(x) = xy
5 x 5 = 25 square units
Physical rectangle?
yes

With x = 5, the constraint gives y = 10 - 5 = 5. Check the perimeter: 2 x 5 + 2 x 5 = 20. The area is 5 x 5 = 25 square units. The 5 by 5 square has the largest area of the three book rectangles: 9, 16, 25.

Use the idea

When a problem has a fixed total, write one quantity in terms of the other first. Then there is only one thing left to choose.

Where the conclusion applies

A rectangle with perimeter exactly 20 units. A real rectangle needs 0 < x < 10. At x = 10 the formula still works but the shape has no width, so the model and the story part ways.

Check your understanding: If x = 3, what are the other side and the area?
y = 10 - 3 = 7, and the area is 3 x 7 = 21 square units.

Chapter 30 source: section "Build the fixed-perimeter rectangle model". Demonstration C30-D01.

Demonstration 2 of 4

The derivative names a candidate

Where does the area stop rising and start falling?

The dashed line touches the curve at the chosen x with slope 10 - 2x. Left of 5 the slope is positive; right of 5 it is negative; at 5 it is zero.

\[A'(x) = 10 - 2x\]

A'(x) is the slope of the area curve at x: square units of area per unit of side. Positive means rising, negative means falling, zero means flat.

Predict first. At x = 6, will the tangent line tilt up or down? What number will A'(6) be?

Choose an example

Figure: The derivative names a candidate. The area curve from 0 to 10 with a dashed tangent line at x = 5, area 25. The tangent has slope 0.
Input x: 5
Constructed example: the chapter's derivative A'(x) = 10 - 2x at the book's inputs 4, 5 and 6, plus 3.

Calculated values

Chosen x
5
A'(x) = 10 - 2x
10 - 2 x 5 = 0
Direction near x
flat (candidate)
Area A(x)
5 x 5 = 25

A'(5) = 10 - 2 x 5 = 10 - 10 = 0. Zero: the tangent is flat. That puts x = 5 on the candidate list. It does not yet prove a maximum; the comparison still has to be made.

Use the idea

A zero derivative tells you where to look. Add that point to a short list, then compare values before calling anything the best.

Where the conclusion applies

The area model A(x) = 10x - x^2, which is smooth everywhere. The slope is a local statement: A'(4) = 2 describes the curve near 4, not across the whole interval.

Check your understanding: What is A'(3), and does area rise or fall there?
A'(3) = 10 - 2 x 3 = 4. It is positive, so area rises as x increases near 3.

Chapter 30 source: section "Derivatives locate interior candidates". Demonstration C30-D02.

Demonstration 3 of 4

The completed square sets a ceiling

Why can no rectangle in this family beat 25 square units?

Every input pays a penalty equal to its squared distance from 5. Inputs the same distance away pay the same penalty, so the curve is symmetric about 5.

\[A(x) = 25 - (x - 5)^2\]

(x - 5)^2 is the squared distance from 5. It is never negative, so 25 minus it is never more than 25.

Predict first. Move 2 units away from 5. Which two inputs share an area, and what is it?

Choose an example

Figure: The completed square sets a ceiling. The curve A(x) = 25 - (x - 5)^2 with a dotted ceiling at 25. Points at x = 4 and x = 6 share the height 24, which is 1 below the ceiling.
Distance from 5: 1
Constructed example: the chapter's pairs 4 and 6, 3 and 7, and its endpoints 0 and 10.

Calculated values

Distance from 5
1
Inputs
4 and 6
Squared distance
1^2 = 1
Area
25 - 1 = 24 square units
Direct check
4 x 6 = 24

A(4) = 25 - (4 - 5)^2 = 25 - 1 = 24, and A(6) = 25 - (6 - 5)^2 = 25 - 1 = 24. Same distance from 5, same area. Direct check: 4 x 6 = 24.

Use the idea

Completing the square is a check that does not use derivatives. When two routes agree on 25 at x = 5, you can trust the answer more.

Where the conclusion applies

The quadratic area model on 0 <= x <= 10. The ceiling argument works because a square is never negative; it would not work for a cubic. At distance 5 the inputs are the degenerate endpoints.

Check your understanding: What area do x = 1 and x = 9 share?
25 - (1 - 5)^2 = 25 - 16 = 9, and 1 x 9 = 9 square units.

Chapter 30 source: section "Complete the square for an independent check". Demonstration C30-D03.

Demonstration 4 of 4

When the turning point is off limits

If the formula peaks at x = 6 but you may not use 6, where is the best point?

The solid part of the curve is the allowed domain. Filled dots are included endpoints, hollow dots are excluded. The star marks the largest attained value, if there is one.

\[B'(x)=12-2x\]

\[B(x)=36-(x-6)^2\]

B(x) = 12x - x^2 is the objective. B'(x) = 12 - 2x is its slope, zero at x = 6. A square bracket includes an endpoint; a round bracket leaves it out.

Predict first. On the open interval (0, 4), is there a largest value at all?

Choose an example

Figure: When the turning point is off limits. The parabola B(x) = 12x - x^2, drawn solid on the domain [0, 4] and dashed elsewhere. Result: maximum 32 at the endpoint x = 4.
Right end of the domain: 4, Endpoints: Included (closed)
Constructed example: the chapter's worked example on [0, 4] and (0, 4), plus wider domains.

Calculated values

Domain
[0, 4]
Is x = 6 allowed?
no
B at the right end
B(4) = 48 - 16 = 32
Largest value
maximum 32 at the endpoint x = 4

B(4) = 36 - (4 - 6)^2 = 36 - 4 = 32. B'(x) = 12 - 2x is zero at 6, outside [0, 4]. The function keeps rising, so the largest allowed value sits at the endpoint 4.

Use the idea

Write the domain beside the objective before differentiating. A turning point outside the allowed inputs is not an answer, however tidy it looks.

Where the conclusion applies

The formula never changes; only the allowed inputs do. On a closed interval the endpoints must be compared. On an open interval the best value can be approached without ever being reached.

Check your understanding: On the closed interval [0, 5], what is the largest value of B(x)?
B' is positive on [0, 5], so the largest is B(5) = 36 - (5 - 6)^2 = 36 - 1 = 35.

Chapter 30 source: section "Worked example: a constraint can exclude the turning point". Demonstration C30-D04.