Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
x = 15 and y = 45
Explanation
This is a constrained optimization problem from the Application of Derivatives chapter. The student must express one variable in terms of the other using the constraint, then maximize the function using derivatives. The key steps involve substitution, differentiation, finding critical points, and verifying the maximum using the second derivative test.
Solution Steps
Step 1: Let the two positive numbers be x and y. Given: x + y = 60, which gives y = 60 - x.
Step 2: Let f(x) = xy³ = x(60 - x)³. We need to maximize this function.
Step 3: Differentiate f(x) using product rule: f'(x) = (60 - x)³ + x · 3(60 - x)² · (-1) = (60 - x)³ - 3x(60 - x)² = (60 - x)²(60 - 4x).
Step 4: Set f'(x) = 0: (60 - x)²(60 - 4x) = 0. This gives x = 60 or x = 15.
Step 5: Find f''(x) = -21600 + 1080x - 12x². At x = 15: f''(15) = -21600 + 16200 - 2700 = -8100 < 0. Hence, x = 15 is a point of local maximum.
Step 6: Since x = 60 gives y = 0 (not positive), the valid solution is x = 15 and y = 60 - 15 = 45.