Unit 4: Applications

Answer: A — The correct answer is A because at the highest point of the trajectory, the ball is momentarily at rest, meaning its velocity is zero. This is due to the acceleration due to gravity, which is constantly acting on the ball, slowing it down as it rises and speeding it up as it falls. The other options are incorrect because the velocity is not increasing (B) or decreasing (C) at the highest point, and it is not constant (D) throughout the trajectory.

Answer: C — To minimize the surface area, we first need to find the value of y when x = 6. Given that the volume V = x^2y = 108 and x = 6, we can substitute x into the equation to get 6^2y = 108. Solving for y, we get 36y = 108, which gives y = 108 / 36 = 3. Therefore, when x = 6, y = 3. The other options do not satisfy the given conditions, so they are incorrect.

Answer: A — The correct answer is A) x = 6, y = 3 because to minimize the cost, we need to minimize the surface area of the box while maintaining the given volume. The volume of the box is given by V = x^2y. Since V = 108, we have x^2y = 108. The cost of the material is given by C = 2x^2 + 4xy. We can express y in terms of x from the volume equation as y = 108/x^2. Substituting this into the cost equation gives C = 2x^2 + 4x(108/x^2) = 2x^2 + 432/x. To minimize the cost, we take the derivative of C with respect to x and set it equal to 0: dC/dx = 4x - 432/x^2 = 0. Solving for x gives x^3 = 108, so x = 6 (since x must be positive). Then y = 108/6^2 = 3. The other options do not minimize the cost.

Answer: B — The correct answer is B. This question tests understanding of related rates—a fundamental application concept. The student must: (1) identify the relationship between variables using V = πr²h, (2) differentiate implicitly with respect to time, recognizing that r is constant but h changes, (3) substitute the known rate dV/dt = -0.5 m³/min and solve for dh/dt. This yields dh/dt = -0.5/(4π) ≈ -0.0398 m/min (negative because water is draining). Why the other answers are wrong: (A) confuses dV/dt with dh/dt and incorrectly divides by radius instead of the cross-sectional area—a common misconception about rates of change. (C) ignores that the relationship between volume change and height change depends on area, not just the radius value. (D) reverses the chain rule relationship; multiplying dV/dh by dV/dt gives dh/dt incorrectly because it should be dV/dt ÷ dV/dh, and also the sign is wrong. This tests the skill of setting up and solving related rates problems, a key application in calculus.

Answer: B — The correct answer is B because to minimize the surface area of the box with a fixed volume and a length twice its width, we need to find the optimal dimensions. Given that length (l) = 2 * width (w) and the volume (V) is 64 cubic meters, we have V = l * w * h. Substituting l = 2w into the volume formula gives us 64 = 2w * w * h, which simplifies to 64 = 2w^2 * h. Since we want to minimize the surface area, A = 2lw + 2lh + 2wh, and knowing l = 2w, we substitute to get A in terms of w and h. However, we also know from the volume equation that h = 64 / (2w^2), so we substitute h in the surface area equation to get A in terms of w only. To minimize A, we take the derivative of A with respect to w, set it equal to zero, and solve for w. This process leads to finding the optimal dimensions that minimize the surface area, which are length = 8m, width = 4m, and height = 2m. The other options do not correctly minimize the surface area given the constraints.