Mathematics 2003 JAMB Past Questions - Question 39: Determine the maximum value of Y= 3X² - x³

Correct Answer

C

Explanation

To determine the maximum value of the function Y = 3X² - X³, we can find the critical points by taking the derivative and setting it equal to zero.Let's find the derivative of Y with respect to X:dY/dX = 6X - 3X²Setting the derivative equal to zero:6X - 3X² = 0Factoring out X:X(6 - 3X) = 0This equation is satisfied when X = 0 or X = 2.To determine if these points are maximum or minimum, we can evaluate the second derivative:d²Y/dX² = 6 - 6XFor X = 0, the second derivative is positive (6 - 6(0) = 6), indicating a minimum point.For X = 2, the second derivative is negative (6 - 6(2) = -6), indicating a maximum point.Therefore, the maximum value of Y = 3X² - X³ occurs when X = 2. We can substitute this value into the equation to find the maximum value:Y = 3(2)² - (2)³Y = 12 - 8Y = 4So, the maximum value of Y is 4.