AP Calculus BC Quiz: Introduction To Optimization Problems

What this quiz covers

This quiz focuses on Introduction To Optimization Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Use $w^2+h^2=100$ and maximize $A=wh$ (correct answer)
2. Use $w+h=10$ and maximize $A=wh$
3. Use $2w+2h=10$ and maximize $A=wh$
4. Use $wh=10$ and minimize $w^2+h^2$
5. Use $w^2+h^2=10$ and maximize $A=wh$

Explanation: This problem involves maximizing the area of a rectangle with a diagonal length constraint. If the rectangle has width w and height h with diagonal fixed at 10, then the constraint is w² + h² = 100 (since diagonal² = w² + h²). We want to maximize the area A = wh subject to this constraint. Choice B incorrectly uses w + h = 10, which would be a perimeter constraint, not a diagonal constraint. The diagonal of a rectangle with sides w and h is √(w² + h²), so fixing the diagonal at 10 gives w² + h² = 100. The approach is to correctly translate geometric constraints (like fixed diagonal length) into algebraic constraint equations.

Question 2

1. Maximize $h(t)$ on the interval $0\le t\le 6$ (correct answer)
2. Minimize $h(t)$ on the interval $0\le t\le 6$
3. Maximize $h'(t)$ on the interval $0\le t\le 6$
4. Minimize $h'(t)$ on the interval $0\le t\le 6$
5. Maximize $t$ on the interval $0\le t\le 6$

Explanation: This problem requires maximizing a quadratic function on a closed interval to find the peak height. The balloon's height is given by h(t) = -t² + 6t + 10, which is a downward-opening parabola. On the interval 0 ≤ t ≤ 6, we want to find the maximum value of h(t). Since this is a standard calculus optimization problem, we maximize h(t) over the given interval. Choice B incorrectly minimizes height, but we want the peak. The vertex occurs at t = -6/(2(-1)) = 3, and h(3) = -9 + 18 + 10 = 19. The fundamental approach is to recognize that finding extreme values (like peak height) requires optimizing the appropriate function over its domain.

Question 3

1. Maximize area $A=lw$ subject to $2l+2w=60$ and $l=2w$ (correct answer)
2. Minimize perimeter $P=2l+2w$ subject to $lw=60$ and $l=2w$
3. Maximize perimeter $P=2l+2w$ subject to $lw=60$ and $l=2w$
4. Minimize area $A=lw$ subject to $2l+2w=60$ and $l=2w$
5. Maximize area $A=2l+2w$ subject to $lw=60$ and $l=2w$

Explanation: This problem involves maximizing the area of a rectangle with both a perimeter constraint and a side ratio constraint. The rectangle has perimeter 2l + 2w = 60 and the constraint l = 2w. Substituting the ratio constraint into the perimeter gives 2(2w) + 2w = 60, so 6w = 60 and w = 10, l = 20. However, the problem asks for the optimization setup before solving. We want to maximize area A = lw subject to 2l + 2w = 60 and l = 2w. Choice B incorrectly minimizes perimeter, but perimeter is fixed. The approach is to set up the area maximization with both the perimeter constraint and the geometric constraint on the side ratio.

Question 4

1. Minimize surface area $S(x)=x^2+4xh$ subject to $x^2h=32$ (correct answer)
2. Maximize surface area $S(x)=x^2+4xh$ subject to $x^2h=32$
3. Minimize volume $V=x^2h$ subject to $x^2+4xh=32$
4. Maximize volume $V=x^2h$ subject to $x^2+4xh=32$
5. Minimize $x$ subject to $x^2+4xh=32$

Explanation: This problem involves minimizing surface area for an open-top box with square base and fixed volume. The box has square base with side x and height h, with volume constraint x²h = 32. The surface area includes the base (x²) and four sides (each xh), giving total S = x² + 4xh. Using the volume constraint h = 32/x², we get S(x) = x² + 4x(32/x²) = x² + 128/x. We minimize this surface area to reduce material costs. Choice B incorrectly maximizes surface area, increasing material usage. The modeling strategy is to use the volume constraint to express surface area as a function of one variable, then optimize to minimize material costs.

Question 5

1. Maximize area $A(x)=x\cdot\frac{16}{x}$ for $x>0$
2. Minimize area $A(x)=x\cdot\frac{16}{x}$ for $x>0$
3. Maximize perimeter $P(x)=2x+2\cdot\frac{16}{x}$ for $x>0$
4. Minimize perimeter $P(x)=2x+2\cdot\frac{16}{x}$ for $x>0$ (correct answer)
5. Maximize $x$ subject to $y=16$

Explanation: This problem requires minimizing the perimeter of a rectangle with vertices constrained by a hyperbola. The rectangle has one vertex at the origin and the opposite vertex at point $(x, \frac{16}{x})$ on the curve $xy = 16$. The rectangle has width $x$ and height $\frac{16}{x}$, giving perimeter $P(x) = 2x + 2(\frac{16}{x}) = 2x + \frac{32}{x}$ for $x > 0$. The area is $A(x) = x(\frac{16}{x}) = 16$, which is constant. Since we want to minimize material used for the frame (perimeter), we minimize $P(x)$. Choice C incorrectly maximizes perimeter, which would waste material. The key insight is that hyperbola constraints create rectangles with constant area, so optimization focuses on minimizing perimeter to reduce material costs.

Question 6

1. Maximize $V(r)=\tfrac13\pi r^2h$ expressed in terms of $r$ using $r^2+h^2=s^2$ (correct answer)
2. Minimize $V(r)=\tfrac13\pi r^2h$ expressed in terms of $r$ using $r^2+h^2=s^2$
3. Maximize $S(r)=\pi r^2+\pi r s$ expressed in terms of $r$ using $V=\tfrac13\pi r^2h$
4. Minimize $S(r)=\pi r^2+\pi r s$ expressed in terms of $r$ using $V=\tfrac13\pi r^2h$
5. Maximize $h$ expressed in terms of $V$ using $s=\tfrac13\pi r^2h$

Explanation: This problem requires maximizing the volume of a cone with fixed slant height constraint. For a cone with base radius $r$, height $h$, and slant height $s$, the relationship is $r^2 + h^2 = s^2$. The volume is $V = (1/3)\pi r^2 h$. With $s$ fixed, we can write $h = \sqrt{s^2 - r^2}$ and substitute to get $V(r) = (1/3)\pi r^2 \sqrt{s^2 - r^2}$. We maximize this volume as a function of $r$, with domain $0 \leq r \leq s$. Choice B incorrectly minimizes volume, but we want the largest possible cup. The key insight is to use the geometric constraint (relating $r$, $h$, and $s$) to express volume as a function of one variable, then optimize that single-variable function.

Question 7

1. The volume of the box as a function of cut size $x$ (correct answer)
2. The surface area of the box as a function of cut size $x$
3. The perimeter of the original sheet as a function of $x$
4. The area of the removed squares as a function of the final volume
5. The height of the box as a function of the sheet’s area

Explanation: This problem involves maximizing the volume of a box created by cutting and folding a flat sheet. When squares of side x are cut from each corner of a rectangular sheet and the sides are folded up, the resulting box has height x. If the original sheet has dimensions L × W, the box has base dimensions (L - 2x) × (W - 2x) and volume V(x) = x(L - 2x)(W - 2x). The domain is 0 < x < min(L/2, W/2). Choice B incorrectly focuses on surface area, but we want maximum storage capacity. The modeling approach is to express the three-dimensional volume in terms of the cut parameter x, accounting for how the cuts affect each dimension of the resulting box.

Question 8

1. Maximize $A=wh + \frac{1}{2} \pi \left( \frac{w}{2} \right)^2$ subject to $w + 2h + \pi \frac{w}{2} = 10$ (correct answer)
2. Minimize $A=wh + \frac{1}{2} \pi \left( \frac{w}{2} \right)^2$ subject to $w + 2h + \pi \frac{w}{2} = 10$
3. Maximize $A=wh$ subject to $2w + 2h = 10$
4. Maximize $A=\pi \left( \frac{w}{2} \right)^2$ subject to $\pi w = 10$
5. Minimize $P=w + 2h$ subject to $wh + \frac{1}{2} \pi \left( \frac{w}{2} \right)^2 = 10$

Explanation: This problem requires maximizing the area of a Norman window subject to a perimeter constraint. A Norman window consists of a rectangle of width $w$ and height $h$ topped by a semicircle of diameter $w$. The total area is $A = wh + \frac{1}{2} \pi \left( \frac{w}{2} \right)^2 = wh + \frac{\pi w^2}{8}$. The perimeter constraint includes the rectangle's base and two sides plus the semicircular arc: $w + 2h + \pi \frac{w}{2} = 10$. Choice C incorrectly treats it as a simple rectangle, ignoring the semicircular top. The modeling strategy is to carefully account for all geometric components in both the objective function and constraint equation.

Question 9

1. Minimize $S(x)=2x^2+4xh$ subject to $x^2h=108$ (correct answer)
2. Maximize $S(x)=2x^2+4xh$ subject to $x^2h=108$
3. Minimize $V(x)=x^2h$ subject to $2x^2+4xh=108$
4. Maximize $V(x)=x^2h$ subject to $2x^2+4xh=108$
5. Minimize $S(x)=x^2h$ subject to $2x^2+4xh=108$

Explanation: This problem involves minimizing surface area for a box with fixed volume constraint. Given that the box has a square base with side x and height h, the volume constraint is x²h = 108. The surface area consists of the base (x²), top (x²), and four sides (each xh), giving total surface area S = 2x² + 4xh. Since we want to minimize material usage, we minimize surface area as a function of x using the volume constraint to eliminate h. Choice B incorrectly suggests maximizing surface area, which would use the most material rather than least. The fundamental approach is to express the objective function in terms of a single variable using the constraint equation.

Question 10

1. The total perimeter as a function of the cut length
2. The total area of the square and triangle as a function of the cut length (correct answer)
3. The area of the square as a function of the triangle’s side length
4. The side length of the square as a function of the triangle’s perimeter
5. The triangle’s perimeter as a function of the square’s area

Explanation: This problem requires minimizing the total area of two shapes formed from a single wire of fixed length. Let x be the length of wire used for the square, so (24 - x) is used for the equilateral triangle. The square has side length x/4 and area (x/4)² = x²/16. The triangle has perimeter (24 - x) and side length (24 - x)/3, giving area (√3/4)((24 - x)/3)². The total area is A(x) = x²/16 + (√3/36)(24 - x)². Choice A incorrectly focuses on perimeter, but total perimeter is fixed at 24 cm. The optimization approach is to express the combined objective (total area) as a function of how the constraint resource (wire length) is allocated between competing uses.

Question 11

1. Maximize revenue $R(x)=x(80-2x)$ (correct answer)
2. Minimize revenue $R(x)=x(80-2x)$
3. Maximize price $p(x)=80-2x$
4. Minimize price $p(x)=80-2x$
5. Maximize demand $x$ subject to $p=80$

Explanation: This problem involves maximizing revenue using a linear demand function. The demand relationship is p = 80 - 2x, where p is price and x is quantity. Revenue is R = xp = x(80 - 2x) = 80x - 2x². This is a quadratic function in x, and we maximize it to find the optimal sales level. The maximum occurs at x = 80/(2·2) = 20, giving maximum revenue R = 80(20) - 2(20)² = 1600 - 800 = 800. Choice B incorrectly minimizes revenue, which would minimize profit. The fundamental approach is to substitute the demand function into the revenue equation and optimize the resulting quadratic function.

Question 12

1. The volume $V=\pi r^2h$ as a function of $r$
2. The lateral area $2\pi rh$ as a function of $h$
3. The total surface area $S=2\pi r^2+2\pi rh$ as a function of $r$ with $\pi r^2h=500$ (correct answer)
4. The height $h$ as a function of $r$ with $S=500$
5. The radius $r$ as a function of $h$ with $S=500$

Explanation: This problem involves minimizing the material needed (surface area) for a cylinder with fixed volume constraint. The cylinder must hold 500 cm³, so the volume constraint is πr²h = 500. The total surface area includes the base (πr²), top (πr²), and lateral surface (2πrh), giving S = 2πr² + 2πrh. To minimize material usage, we minimize this surface area as a function of r, using the volume constraint to express h in terms of r. Choice A incorrectly suggests maximizing volume, but volume is fixed by the problem requirement. The optimization approach is to minimize cost-related quantities (here, material surface area) subject to performance constraints (required volume).

Question 13

1. The rectangle’s area as a function of the rightmost $x$-coordinate (correct answer)
2. The rectangle’s perimeter as a function of the parabola’s vertex
3. The parabola’s maximum value as a function of the rectangle’s width
4. The rectangle’s height as a function of its perimeter
5. The $x$-intercepts as a function of the rectangle’s area

Explanation: This problem involves maximizing the area of a rectangle with vertices constrained by a parabola and the x-axis. The rectangle has one side along the x-axis and two vertices on the parabola y = 4x - x². If the rectangle extends from x = 0 to x = a, then its width is a and height is 4a - a², giving area A(a) = a(4a - a²) = 4a² - a³. The domain is 0 ≤ a ≤ 4 since the height must be non-negative. Choice B incorrectly focuses on perimeter, but we want maximum enclosed area. The modeling strategy is to parameterize the rectangle using one coordinate (the rightmost x-coordinate), then use the parabola equation to determine the height and express area as a single-variable function.

Question 14

1. Maximize $A=xy$ with constraint $3x+2y=12$ (correct answer)
2. Minimize $A=xy$ with constraint $3x+2y=12$
3. Maximize $P=2x+2y$ with constraint $xy=12$
4. Minimize $P=2x+2y$ with constraint $xy=12$
5. Maximize $3x+2y$ with constraint $xy=12$

Explanation: This problem requires maximizing the area of a rectangle with vertices on coordinate axes and a line. The rectangle has vertices at (0,0), (x,0), (0,y), and (x,y) where the point (x,y) lies on the line 3x + 2y = 12 in the first quadrant. The rectangle's area is A = xy, and we maximize this subject to the line constraint. Using the constraint y = (12 - 3x)/2, we get A(x) = x(12 - 3x)/2 = (12x - 3x²)/2. This is maximized when x = 2, giving y = 3 and maximum area 6. Choice B incorrectly minimizes area, but we want maximum enclosed space. The fundamental approach is to use the line constraint to express area as a function of one coordinate, then optimize that function.

Question 15

1. Minimize $D(x)=\sqrt{(x-2)^2+25}$ for all real $x$ (correct answer)
2. Maximize $D(x)=\sqrt{(x-2)^2+25}$ for all real $x$
3. Minimize $D(x)=\sqrt{(x+2)^2+25}$ for $x\ge 0$
4. Minimize $D(x)=(x-2)^2+25$ for $x\ge 5$
5. Maximize $D(x)=(x-2)^2+25$ for all real $x$

Explanation: This problem requires minimizing the distance from a fixed point to a variable point constrained to lie on the x-axis. The fixed point is (2,5) and the variable point is (x,0). The distance is D(x) = √[(x-2)² + (0-5)²] = √[(x-2)² + 25]. Since we want to minimize this distance, and x can be any real number, we minimize D(x) = √[(x-2)² + 25]. Choice C incorrectly uses (x+2)² instead of (x-2)², and choice D unnecessarily restricts the domain. The fundamental approach is to use the distance formula and recognize that minimizing distance is equivalent to minimizing the squared distance when the square root complicates differentiation.

Question 16

1. Maximize $A=lw$ subject to the constraint that the diagonal satisfies $l^2+w^2=100$ (correct answer)
2. Minimize $A=lw$ subject to the constraint that the diagonal satisfies $l^2+w^2=100$
3. Maximize $P=2l+2w$ subject to $lw=25$
4. Minimize $P=2l+2w$ subject to $lw=25$
5. Maximize $l+w$ subject to $l^2+w^2=25$

Explanation: This problem involves maximizing the area of a rectangle inscribed in a circle. For a rectangle inscribed in a circle of radius 5, the diagonal of the rectangle equals the diameter of the circle, which is 10. If the rectangle has length l and width w, then l² + w² = 100 (since diagonal = 10). We want to maximize the area A = lw subject to this constraint. Choice B incorrectly minimizes area, but we want the largest possible rectangle. Using the method of Lagrange multipliers or substitution, the maximum occurs when l = w = 5√2, giving maximum area 50. The approach is to use the geometric constraint (rectangle inscribed in circle means diagonal equals diameter) to set up the optimization problem.

Question 17

1. The area of the rectangular pen as a function of the two fenced side lengths (correct answer)
2. The perimeter of the pen as a function of the two fenced side lengths
3. The sum of the three fenced side lengths as a function of the unfenced side length
4. The length of fencing used as a function of the pen’s area
5. The area of the pen as a function of the total fencing available, 200

Explanation: This problem requires setting up an optimization to maximize enclosed area with a constraint on available fencing. The farmer wants to create the largest possible rectangular pen using 200 m of fencing, where one side borders a river and needs no fence. If the pen has width w (perpendicular to river) and length l (parallel to river), then the constraint is w + w + l = 200, or 2w + l = 200. The objective is to maximize the area A = wl as a function of the fence dimensions. Choice B incorrectly suggests maximizing perimeter, but perimeter is fixed by the constraint. The key insight is that optimization problems maximize or minimize an objective function (here, area) subject to constraint equations involving the decision variables.

Question 18

1. Minimize total travel time as a function of $x$ (correct answer)
2. Maximize total travel time as a function of $x$
3. Minimize total distance as a function of $x$
4. Maximize total distance as a function of $x$
5. Minimize $x$ as a function of the final point

Explanation: This problem involves minimizing total travel time when speeds differ on different segments of a path. The runner travels from (0,0) to point (x,0) at one speed, then from (x,0) to (10,6) at a different speed. If the speeds are v₁ and v₂ respectively, the total time is T(x) = x/v₁ + √[(10-x)² + 36]/v₂. The first distance is x, and the second distance is √[(10-x)² + 6²] by the distance formula. We minimize total travel time as a function of the intermediate point x. Choice B incorrectly maximizes time, seeking the slowest route. The fundamental approach is to express total time as the sum of segment times, each calculated as distance divided by speed for that segment.

Question 19

1. The rectangle’s area as a function of $x$ using $y=\sqrt{r^2-(x/2)^2}$ (correct answer)
2. The semicircle’s area as a function of $x$
3. The rectangle’s perimeter as a function of $r$
4. The rectangle’s height as a function of its area
5. The radius $r$ as a function of the rectangle’s width

Explanation: This problem involves maximizing the area of a rectangle inscribed in a semicircle with geometric constraints. The rectangle has width x and is symmetric about the y-axis, so it extends from -x/2 to x/2 horizontally. The height is determined by the semicircle equation: x² + y² = r², so y = √[r² - (x/2)²]. The rectangle's area is A(x) = x · √[r² - (x/2)²]. The domain is 0 ≤ x ≤ 2r. Choice B incorrectly focuses on the semicircle's area, which is fixed. The modeling strategy is to use one parameter (width) and the constraint curve to express the area as a single-variable function.

Question 20

1. Maximize area $A(x)=x(10-x)$ for $0\le x\le 10$ (correct answer)
2. Minimize area $A(x)=x(10-x)$ for $0\le x\le 10$
3. Maximize perimeter $P(x)=x+(10-x)$ for $0\le x\le 10$
4. Minimize perimeter $P(x)=2x+2(10-x)$ for $0\le x\le 10$
5. Maximize height $h(x)=x$ for $0\le x\le 10$

Explanation: This problem requires maximizing the area of a rectangle with vertices constrained by a line and the axes. The rectangle has one side on the x-axis from (0,0) to (x,0) and extends up to the line y = 10 - x. The rectangle's width is x and height is 10 - x, giving area A(x) = x(10 - x) = 10x - x². The domain is 0 ≤ x ≤ 10 since both width and height must be non-negative. Choice B incorrectly minimizes area, but we want maximum enclosed space. The key insight is to parameterize the rectangle using one coordinate, then use the constraint (the line equation) to express the area as a single-variable function.