The annual total revenue for a product is given by R(x)=63,000x−9x2 dollars, where x is the number of units sold.?

Method 1 : Completing square

R(x) = 63,000x - 9x²
R(x) = -9(x² - 7,000)
R(x) = -9(x² - 7,000 + 3,500²) + 9*3,500²
R(x) = -9(x - 3,500)² + 110,250,000

For all real values of x, -9(x - 3,500)² ≤ 0
Hence, R(x) = -9(x - 3,500)² + 110,250,000 ≤ 110,250,000
Maximum R(x) = 110,250,000 when x = 3500

To maximize revenue, __3500__ units must be sold.
The maximum possible annual revenue is __110,250,000__.

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Method 2 : Differentiation

R(x) = 63,000x - 9x²
R'(x) = 63,000 - 18x = -18(x - 3,500)
R"(x) = -18

When x = 3500 :
R'(x) = 0
R"(x) = -18 < 0
Hence, maximum R(x) when x = 3,500
Maximum R(x) = 63,000(3,500) - 9(3500)² = 110,250,000

To maximize revenue, __3500__ units must be sold.
The maximum possible annual revenue is __110,250,000__.

Vertex of a parabola is at x = -b/(2a).
x = -63,000 / (2 * -9)
x = 3500

R(3500) = ???