- A lifeguard runs then swims to a victim; total time depends on run distance x. What should be minimized?
Answer: Minimize total travel time T(x) (run time plus swim time).
This classic lifeguard problem requires minimizing rescue time through optimal path selection. The lifeguard runs distance x along the beach (at running speed) then swims the remaining diagonal distance to the victim (at swimming speed). The total time T(x) = (run time) + (swim time) depends on where the lifeguard enters the water. We minimize T(x) to reach the victim fastest. Option D incorrectly suggests maximizing time, which would delay rescue. The key insight is that the fastest path balances the speed advantage of running against the shorter diagonal swim distance. - A swimmer starts 3 km downshore from a dock and can swim 2 km/h, run 6 km/h. What should be minimized?
Answer: The total travel time (swimming time plus running time)
This question tests the skill of setting up optimization problems by identifying the quantity to be optimized in a path minimization scenario. The swimmer starts 3 km downshore from the dock, swimming at 2 km/h and running at 6 km/h, so the objective is to minimize the total travel time, which is swimming time plus running time. The path involves swimming to some point on the shore and then running to the dock, with time depending on distances and speeds. We model the total time as a function of the landing point to find the minimum. A tempting distractor is choice A, the total distance traveled, but minimizing distance ignores different speeds and does not minimize time. In optimization problems involving different media or speeds, express time as the objective function and minimize it with respect to the path variable.
- A river is 50 m wide; a cable runs from point A to point B on opposite banks. What is optimized to save cable?
Answer: Minimize the cable length between A and B.
This cable routing problem minimizes material usage across a river. To connect points A and B on opposite banks of a 50 m wide river, we need to find the path requiring the least cable length. This involves choosing where to cross the river to minimize total distance. Option A incorrectly suggests maximizing cable length, which would waste material and increase cost. The optimization principle for routing problems is minimizing total distance (or cost) while satisfying connectivity requirements.
- A rectangular garden uses 80 m of edging on all four sides. What objective quantity should be maximized?
Answer: The garden's area
This question tests the skill of setting up optimization problems by identifying the quantity to be maximized with a fixed perimeter for a garden. The rectangular garden uses 80 m of edging on all four sides, so the objective is to maximize the garden's area to get the most space for planting. The perimeter constraint is 2(length + width) = 80, allowing us to express area as length times width in terms of one variable. This setup comes from efficiently using limited edging to enclose the largest possible area. A tempting distractor is choice B, the garden's perimeter, but this is fixed at 80 m and not what we maximize. In optimization problems with fixed perimeters, model the area as the objective function and use the perimeter constraint to reduce variables for finding maxima.
- A Norman window is a rectangle topped by a semicircle with fixed perimeter 12 m; what objective is optimized?
Answer: Maximize total area of the window, A, subject to perimeter 12 m.
This problem optimizes a Norman window design with fixed perimeter. A Norman window consists of a rectangle (width w, height h) topped by a semicircle (radius r = w/2), giving total perimeter P = 2h + w + πw/2 = 12 m. To maximize natural light, we want to maximize the total area A = wh + πw²/8. Option E incorrectly suggests maximizing perimeter when it's already constrained at 12 m. The optimization principle here is maximizing benefit (window area for light) while respecting the constraint (fixed frame material). - A right circular cylinder has fixed volume V=πr2h=1000; which quantity should be minimized to reduce total surface area?
Answer: The total surface area, S=2πr2+2πrh
This problem asks what to minimize when reducing the total surface area of a cylinder with fixed volume. A cylinder's total surface area includes two circular ends (each with area πr²) and the lateral surface (area 2πrh), giving S = 2πr² + 2πrh. Since volume V = πr²h = 1000 is fixed, we can express h = 1000/(πr²) and substitute into the surface area formula to get S as a function of r alone. Choice D (lateral area only) is incorrect because minimizing just the lateral surface ignores the material needed for the two circular ends. In surface area minimization problems, include all surfaces that require material—don't forget tops, bottoms, or any face of the three-dimensional object. - A company sells q items; revenue is R(q)=qp(q) with price p(q)=50−0.2q; what should be maximized?
Answer: The revenue, R(q)=q(50−0.2q)
This problem involves maximizing revenue in a pricing optimization scenario. Revenue equals the number of items sold (q) times the price per item (p(q)), giving R(q) = q · p(q) = q(50 - 0.2q) = 50q - 0.2q². The company wants to find the quantity q that maximizes this revenue function. As quantity increases, price decreases linearly, creating a trade-off between selling more items at lower prices versus fewer items at higher prices. Choice A (the price p(q)) is tempting but incorrect because maximizing price alone would mean selling zero items at the highest possible price, yielding zero revenue. In business optimization, revenue maximization requires balancing price and quantity, not maximizing either variable independently. - A poster has printed area xy and requires 1-inch margins on all sides; which quantity should be minimized for least paper used?
Answer: The total paper area, (x+2)(y+2)
This problem asks what to minimize when reducing paper usage for a poster with margins. If the printed area has dimensions x by y, then with 1-inch margins on all sides, the paper dimensions are (x + 2) by (y + 2), giving total paper area (x + 2)(y + 2). To minimize paper usage while maintaining a specific printed area xy = constant, we minimize the total paper area. The margin constraint is built into the relationship between printed and paper dimensions. Choice A (printed area) is incorrect because the printed area is typically fixed by content requirements, not something we minimize. When dealing with margin problems, the objective is to minimize the total material (paper) area while maintaining the required usable (printed) area. - A box with square base and no lid has volume 32 ft3. To minimize cost, what should be minimized?
Answer: The total surface area of cardboard used (base plus four sides)
This question tests the skill of setting up optimization problems by identifying the quantity to be minimized for a box with fixed volume. The box has a square base and no lid, with a volume of 32 ft³, and to minimize cost, we minimize the total surface area of the cardboard used, which includes the base and four sides. The cost is assumed proportional to the surface area, so we relate the side length of the base and height through the volume constraint. Expressing the surface area as a function of one variable allows us to find the dimensions that use the least material. A tempting distractor is choice A, the volume of the box, but this is fixed at 32 ft³ and not what we minimize. In optimization problems for open containers with fixed volume, always model the surface area as the objective function using the volume constraint to minimize material costs.
- A Norman window is a rectangle topped by a semicircle; if the perimeter is fixed, what should be maximized?
Answer: The window's total area (rectangle plus semicircle)
This problem involves maximizing the area of a Norman window (rectangle topped by semicircle) with fixed perimeter. If the rectangle has width 2r and height h, and the semicircle has radius r, then the total area is A = 2rh + (πr²/2). The perimeter constraint includes the three sides of the rectangle plus the semicircular arc: 2h + 2r + πr = P (fixed). Using this constraint to express h in terms of r allows us to maximize A as a function of r alone. Choice B (perimeter) is incorrect because perimeter is the given constraint, not what we're optimizing. In composite shape problems, identify all components contributing to both the objective function and constraints.
- A box with square base and no lid must hold 108 cm3; what should be minimized in setup?
Answer: Minimize surface area, S=x2+4xh, subject to x2h=108.
This optimization problem involves minimizing material usage for a box with fixed volume. The box has a square base (side length x) and height h, with volume constraint x²h = 108 cm³. Since the box has no lid, the surface area is S = x² + 4xh (one base plus four sides). To use the least material, we minimize this surface area subject to the volume constraint. Option A incorrectly suggests maximizing surface area, which would waste material. When dealing with containers, minimizing surface area for fixed volume is a common cost-saving objective. - A company sells x items; profit is P(x)=R(x)−C(x). To optimize profit, what should be maximized?
Answer: The profit P(x)
This question tests the skill of setting up optimization problems by identifying the quantity to be optimized in a business context. The company sells x items, with profit defined as P(x)=R(x)−C(x), where R is revenue and C is cost, so the objective is to maximize the profit function. This arises from the need to find the production level x that yields the highest net gain after subtracting costs from revenue. By modeling profit as a function of x, we can use calculus to locate the maximum. A tempting distractor is choice A, the cost C(x), but minimizing cost alone ignores revenue and does not optimize profit. In optimization problems involving profit, express profit as revenue minus cost and maximize it with respect to the variable like quantity sold. - A farmer has 200 m of fencing to enclose a rectangular pen; what quantity should be maximized to use calculus?
Answer: Maximize the area of the rectangle, A=LW, subject to 2L+2W=200.
This problem asks us to set up an optimization scenario for maximizing the enclosed area with fixed fencing. The farmer has a constraint of 200 m of fencing, which gives us the perimeter equation 2L + 2W = 200. Since we want the largest possible pen, we need to maximize the area A = LW subject to this perimeter constraint. Option B incorrectly suggests maximizing perimeter when it's already fixed at 200 m. The key strategy is to identify what quantity benefits the farmer (area) versus what resource is limited (perimeter). - A right circular cone has fixed slant height 10 cm; to hold most volume, what should be maximized?
Answer: Maximize volume, V=31πr2h, subject to r2+h2=100.
This cone optimization problem maximizes volume with fixed slant height. A right circular cone with slant height ℓ = 10 cm satisfies r² + h² = 100 (Pythagorean theorem). To hold maximum volume, we maximize V = (1/3)πr²h subject to this constraint. Option C incorrectly suggests minimizing volume, which would create an inefficient container. For three-dimensional containers with geometric constraints, we typically maximize volume (capacity) or minimize surface area (material cost). - A rectangle is inscribed under y=9−x2 in the first quadrant with one corner at the origin; what should be maximized?
Answer: The rectangle's area, A=xy with y=9−x2
This problem involves maximizing the area of a rectangle inscribed under a parabola. With one corner at the origin and the opposite corner at (x, y) on the parabola y = 9 - x², the rectangle has width x and height y = 9 - x². Therefore, the area to maximize is A = xy = x(9 - x²) = 9x - x³. The constraint that the rectangle fits under the parabola is automatically satisfied by using y = 9 - x² for the height. Choice E (perimeter) is incorrect because we want to maximize usable area, not the distance around the rectangle's edge. When inscribing rectangles under curves, express the rectangle's dimensions using the curve's equation to ensure the geometric constraint is satisfied. - A rectangular poster has area 600 cm2 and 2 cm margins on all sides. What should be maximized?
Answer: The printed (usable) area inside the margins
This question tests the skill of setting up optimization problems by identifying the quantity to be optimized for a poster with fixed total area and margins. The poster has a total area of 600 cm², and with 2 cm margins on all sides, the goal is to maximize the printed usable area inside these margins by choosing optimal dimensions. The printed area is (length - 4 cm) times (width - 4 cm), and we use the fixed total area constraint to express this in one variable. This setup arises from wanting the largest possible printable space on a sheet of fixed size while maintaining uniform margins. A tempting distractor is choice B, the total poster area, but this is fixed at 600 cm² and not what we maximize. In optimization problems with fixed total resources and subtractive elements like margins, express the usable quantity as the objective and optimize dimensions accordingly.
- A rectangular garden uses 40 m of fencing for three sides and a wall for the fourth; what should be maximized?
Answer: The garden's area, A=xy
This problem requires identifying the optimization objective when building a rectangular garden with limited fencing. The garden uses 40 m of fencing for three sides (two widths and one length), while the fourth side is against an existing wall. Since we want to maximize the usable space with our limited resources, the objective is to maximize the garden's area A = xy. The constraint that only three sides need fencing (2x + y = 40) allows us to express y in terms of x and create a single-variable optimization problem. Choice C incorrectly includes all four sides in the perimeter calculation, failing to account for the wall providing the fourth side. When facing optimization problems, always identify what quantity you want to maximize or minimize based on the practical goal of the situation. - A company sells x items; profit is P(x)=R(x)−C(x). For optimization, what should be maximized?
Answer: Maximize profit P(x) over feasible x.
This business optimization problem seeks to maximize profit from selling items. Profit is defined as P(x) = R(x) - C(x), where R(x) is revenue and C(x) is cost for producing/selling x items. To find the optimal production level, we maximize P(x) over feasible values of x. Option D incorrectly suggests minimizing profit, which would harm the business. The fundamental principle in business optimization is maximizing profit (or minimizing loss) subject to practical constraints. - A right circular cone has fixed slant height 10 cm. To hold the most, what should be maximized?
Answer: The cone's volume 31πr2h
This question tests the skill of setting up optimization problems by identifying the quantity to be maximized for a cone with fixed slant height. The right circular cone has a fixed slant height of 10 cm, and to hold the most volume, we maximize the volume given by (1/3)π r² h. The slant height relates radius r and height h via the Pythagorean theorem, serving as the constraint. This allows expressing volume in terms of one variable to find the maximum. A tempting distractor is choice A, the lateral surface area π r ℓ, but this is not the quantity we aim to optimize for capacity. In optimization problems with geometric constraints like fixed slant height, model the volume as the objective and use the constraint to relate dimensions for maximum capacity. - A point on y=9−x2 in the first quadrant forms a rectangle with axes; what quantity should be maximized?
Answer: Maximize the rectangle's area, A=x⋅y, with y=9−x2.
This geometric optimization problem maximizes the area of a rectangle formed by a point on a parabola. A point (x, y) on the parabola y = 9 - x² in the first quadrant forms a rectangle with the coordinate axes having area A = xy = x(9 - x²). We maximize this area function to find the optimal point location. Option B incorrectly suggests minimizing area, which would give a degenerate rectangle. The strategy is to express the objective (rectangle area) in terms of a single variable using the constraint (point on parabola).
