Math question: What is the maximum value and when does it occur?

Method 1 : By completing square

h(x) = -4500x² + 36000x + 138000
h(x) = -4500 (x² - 8x) + 138000
h(x) = -4500 (x² - 8x + 4²) + 4500*4² + 138000
h(x) = 210000 - 4500 (x - 4)²

For all real values of x, (x - 4)² ≥ 0 and thus -4500 (x - 4)² ≤ 0
Hence, h(x) = 210000 - 4500 (x - 4)² ≤ 210000
Maximum h(x) = 210000 when x = 4

Maximum value = $210000 when time = 4 years

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Method 2 : Use differentiation
h(x) = -4500x² + 36000x + 138000
h'(x) = -9000x + 36000
h"(x) = -9000

When x = 4, h'(x) = 0 and h"(x) < 0
When x = 4, maximum h(x) = -4500(4)² + 36000(4) + 138000 = 210000

Maximum value = $210000 when time = 4 years

A local max or min happens when

dh(x)/dx = 0 so let's do the math

dh(x)/dx = -9000 x + 36000 = 0 solve for x

x = 36000/9000 = 4

Check to see if max or min by computing d^2h(x)/dx^2, evaluate at x = 4 and look at the sign.

d^2h(x)/dx = -9000 --> since this is < 0 it means the function has a local (and in this case global) maximum at x = 4

Now compute h(4) = -4500*(4)^2 + 36000*(4) + 138000 = 192,000