This problem involves inscribing a rectangle under a parabola, requiring us to identify the optimization objective. With the parabola y = 16 - x² and rectangle corners at (±x, 0) and (±x, 16-x²), the rectangle has width 2x and height 16-x². The area is A(x) = 2x(16-x²), which should be maximized to find the largest possible inscribed rectangle. The perimeter (choice D) would give P = 4x + 2(16-x²), but maximizing perimeter doesn't necessarily give the largest area. For inscribed figure problems, maximize area unless specifically asked for another quantity.

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.

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.

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.

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.

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.

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.

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).

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.

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.