Question

Question 3: Math Review-(Unconstrained) Optimization To find the value of x that maximizes f(x), first find the values of z that satisfies f,(z) 0 and f"(x) <0. If there is only one such value, that is the value of r that maximizes f(x). If there are many values of z, say x1,T2, ,ZA, that satisfy those conditions, compare the values f(x), f(r2),..../(xN) and find out which r among i, r2,... .xv gives the global maximum ชา 1. Find out the value of a that maximizes f(a)-2 2. 2. Find out the value of r that maximizes f(x)t -3 + 22.

Answer 1

(1)

f(x) = -x² + 2x

f'(x) = -2x + 2 = 0

2x = 2

x = 1

Since f''(x) = -2 < 0, this is a maxima.

(2)

f(x) = -(x⁴/4) + (x³/3) + x² + 2

f'(x) = -x³ + x² + 2x = 0

x(-x² + x + 2) = 0

Either x = 0, or (-x² + x + 2) = 0

If (-x² + x + 2) = 0,

x² - x - 2 = 0

x² + x - 2x - 2 = 0

x(x + 1) - 2(x + 1) = 0

(x + 1) (x - 2) = 0

x = -1 or x = 2

Therefore, possible values are: x = -1, x = 0 or x = 2.

When x = -1, f(x) = (-1/4) - (1/3) + 1 + 2 = (- 3 - 4 + 12 + 24) / 12 = 29/12 = 2.42

When x = 0, f(x) = -0 + 0 + 0 + 2 = 2

When x = 2, f(x) = (-16/4) + (8/3) + 4 + 2 = -4 + (8/3) + 6 = 2 + (8/3) = (6 + 8) / 3 = 14/3 = 4.67

Therefore, f(x) is maximized when x = 2.