To maximize f(x) = −6x + 3x^2 − 2x^3, subject to x ≥ 0, we need to find the critical points of f(x) where the first derivative is zero or undefined, and then use the second derivative test to determine if they are local maxima or minima.
The first derivative of f(x) is f’(x) = −6 + 6x − 6x^2. To find the critical points, we set f’(x) equal to zero and solve for x:
f’(x) = 0 −6 + 6x − 6x^2 = 0 −1 + x − x^2 = 0 x^2 − x − 1 = 0 (x − (1 + √5)/2)(x − (1 − √5)/2) = 0 x = (1 + √5)/2 or x = (1 − √5)/2
Since x ≥ 0, we only consider the positive root, x = (1 + √5)/2 ≈ 1.618.
The second derivative of f(x) is f’‘(x) = 6 − 12x. To use the second derivative test, we plug in the critical point x = (1 + √5)/2 and see if f’'(x) is positive or negative:
f’'((1 + √5)/2) = 6 − 12(1 + √5)/2 = 6 − 6(1 + √5) = −6√5 < 0
Since f’'(x) is negative at the critical point, it means that f(x) is concave down and has a local maximum there.
Therefore, the optimal solution is to choose x = (1 + √5)/2, which gives the maximum value of f(x) = −6(1 + √5)/2 + 3(1 + √5)^2/4 − 2(1 + √5)^3/8 ≈ 0.541.
I hope this helps you solve the problem. Have a good day!