# The revenue from the sale of a product is given by R=1642x-11x^2 - x^3.?

If the sale of 11 units gives a total revenue of 15400\$, find another number of units that will give 15400\$ in revenue.

Relevance
• You are given:

R = 1642x - 11x² - x³

We are interested in values that make R = 15400, so let's set that in place and put it into standard form:

15400 = 1642x - 11x² - x³

0 = 1642x - 11x² - x³ - 15400

0 = -x³ - 11x² + 1642x - 15400

Let's multiply both sides by -1 to simplify as I don't like having a negative leading coefficient if I can help it:

0 = x³ + 11x² - 1642x + 15400

We are told that x = 11 is a solution to this quadratic so (x - 11) must be a factor. We can divide the cubic by the linear expression to get a quadratic in result:

. . . . .__x²_+_22x_-_1400__________

x - 11 ) x³ + 11x² - 1642x + 15400

. . . . . .x³ - 11x²

. . . . . ---------------

. . . . . . . . . 22x² - 1642x + 15400

. . . . . . . . . 22x² - 242x

. . . . . . . . ------------------------

. . . . . . . . . . . . . -1400x + 15400

. . . . . . . . . . . . . -1400x + 15400

. . . . . . . . . . . . .-------------------------

. . . . . . . . . . . . . . . . . . . . . . . 0

Now we have this quadratic to be solved:

x² + 22x - 1400 = 0

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -22 ± √(22² - 4(1)(-1400))] / (2 * 1)

x = [ -22 ± √(484 + 5600)] / 2

x = [ -22 ± √(6084)] / 2

x = (-22 ± 78) / 2

x = -100/2 and 56/2

x = -50 and 28

We aren't interested in the negative value, so the other solution that question is asking for is:

28 units to make the same \$15,400 in revenue as the 11 units we were given.

• R = 1642x - 11x² - x³

first check that 11 gives you \$15400

R = 1642•11 - 11•121 - 1331 = 15400

ok

1642x - 11x² - x³ = 15400

solve for x

graphically, the + solutions are 11 and 28