AP Calculus AB Optimization Test Solutions and Scoring (PDF)

AP Calculus AB math test solutions: Optimization – Maths PDF

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This optimization test answers for AP Calculus AB covers optimization (Common Core AP CALC AB 5.10, AP CALC AB 5.11), with a free PDF to print.

Use this answer key to score the optimization test out of 20 points. Each solution shows the work and a scoring guide that says how the points are awarded. Give partial credit for a correct setup, a correct derivative or integral, and a correct final statement with units where they apply.

AP Calculus AB · Common Core State Standards: AP CALC AB 5.10, AP CALC AB 5.11

Optimization test answers: worked solutions

Question 1 – Maximum product [2 pts]

$P=x(30-x)$, $P^{\prime}=30-2x=0$ gives $x=15$. $P^{\prime\prime}=-2 < 0$. Both numbers are $15$ (product $225$).

Scoring: 1 point for the function and derivative; 1 point for the answer with justification.

Question 2 – Pen against a wall [3 pts]

$2x+y=80$, $y=80-2x$. $A=x(80-2x)=80x-2x^{2}$, $0 < x < 40$.

$A^{\prime}=80-4x=0$ gives $x=20$, $y=40$. $A^{\prime\prime}=-4 < 0$.

Final answer: 20 ft by 40 ft; maximum area 800 ft².

Pen against a wall; 80 ft of fence.

Scoring: 1 point for the area function with domain; 1 point for the critical number; 1 point for the justification and final answer.

Question 3 – Open box [3 pts]

$V=x(18-2x)^{2}$, $0 < x < 9$. $V^{\prime}=(18-2x)^{2}-4x(18-2x)=(18-2x)(18-6x)$.

Critical numbers: $x=9$ (endpoint) and $x=3$. $V^{\prime}$ goes from $+$ to $-$ at $x=3$. $V(3)=3\cdot 144=432$.

Final answer: $x=3$ in, maximum volume 432 in³.

An 18-inch square sheet.

Scoring: 1 point for V(x); 1 point for the derivative and critical number; 1 point for justification and the volume.

Question 4 – Closest point [2 pts]

$D^{2}=x^{2}+(x+3)^{2}=2x^{2}+6x+9$. $4x+6=0$ gives $x=-\frac{3}{2}$, $y=\frac{3}{2}$.

Final answer: $\left(-\frac{3}{2},\frac{3}{2}\right)$.

Scoring: 1 point for the distance function; 1 point for the point.

Question 5 – Revenue [3 pts]

$R(p)=p(300-6p)=300p-6p^{2}$.
$R^{\prime}=300-12p=0$ gives $p=25$.
$R^{\prime\prime}=-12 < 0$, a maximum; $R(25)=25\cdot 150=3750$ dollars per week.

Scoring: 1 point per part.

Question 6 – Can design [3 pts]

$h=\frac{1000}{\pi r^{2}}$. $S=2\pi r^{2}+2\pi rh=2\pi r^{2}+\frac{2000}{r}$.

$S^{\prime}=4\pi r-\frac{2000}{r^{2}}=0$ gives $r^{3}=\frac{500}{\pi}$, $r\approx 5.42$ cm. $S^{\prime\prime}=4\pi+\frac{4000}{r^{3}} > 0$: minimum.

A can holding 1000 cm³.

Scoring: 1 point for S(r); 1 point for the critical number; 1 point for the justification.

Question 7 – Justifying an extremum [2 pts]

$A^{\prime}=60-4x=0$ gives $x=15$. $A^{\prime} > 0$ for $0 < x < 15$ and $A^{\prime} < 0$ for $15 < x < 30$.

So $A$ increases then decreases: absolute maximum $A(15)=15\cdot 30=450$.

Scoring: 1 point for the sign analysis; 1 point for the conclusion and value.

Question 8 – Ticket price [2 pts]

With $n$ increases of \$2: price $20+2n$ and tickets $400-20n$. $R(n)=(20+2n)(400-20n)=8000+400n-40n^{2}$.

$R^{\prime}=400-80n=0$ gives $n=5$. Price $20+10=30$, revenue $30\cdot 300=9000$.

Final answer: \$30 per ticket, revenue \$9000.

Scoring: 1 point for R(n); 1 point for the answer.

Total score

Total

20 points

Key facts behind this optimization test answers

Key facts

Name the variables, write an objective function, and use a constraint to reduce it to one variable.
Find critical numbers by solving f′ = 0 on the allowed domain.
Justify the extreme with a sign change of f′, the sign of f″, or by comparing endpoint values.
Check the domain and give the final answer in context with units.

How to use this optimization test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

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