The research department in a company that manufactures AM/FM clock radios established the following cost, and revenue functions:

R(x) = x(50-1.25x)
C(x) = 160 + 10x
where 0 < x < 20.

a) Determine when R = C (to the nearest thousand units. e.g. 6.247)
Quantity x = ________ thousand units.

b) Determine the maximum profit (to the nearest thousand dollars).
Profit P = _______ thosand dollars.

My answers don't seem to be correct (x = 16 and P = 160) Please help!

Is max profit 80 than? I am plugging it back into P(x) = = 50x - 1.25x^2 - 160 - 10x right?

yes!, good job

To determine the quantity x when R = C, you need to set the revenue function R(x) equal to the cost function C(x) and solve for x:

R(x) = C(x)
x(50 - 1.25x) = 160 + 10x

To solve this equation, first distribute x to both terms in the parentheses on the left side:

50x - 1.25x^2 = 160 + 10x

Combine like terms by moving all terms to one side of the equation:

1.25x^2 - 40x + 160 = 0

To solve this quadratic equation, you can either factor it or use the quadratic formula. Factoring seems difficult in this case, so we'll use the quadratic formula:

x = [-(-40) ± √((-40)^2 - 4(1.25)(160))] / (2 * 1.25)

Simplifying further:

x = (40 ± √(1600 - 800)) / 2.5
x = (40 ± √(800)) / 2.5
x = (40 ± 28.284) / 2.5

Now, solve for both possible values of x:

x1 = (40 + 28.284) / 2.5
x1 = 68.284 / 2.5
x1 ≈ 27.314

x2 = (40 - 28.284) / 2.5
x2 = 11.716 / 2.5
x2 ≈ 4.686

Since the given range is 0 < x < 20, we disregard x1 ≈ 27.314 and consider the solution x2 ≈ 4.686.

Therefore, the quantity x when R = C is approximately 4.686 thousand units.

To determine the maximum profit, you need to find the value of x that maximizes the profit function P(x), which is calculated by subtracting the cost function C(x) from the revenue function R(x):

P(x) = R(x) - C(x)
P(x) = x(50 - 1.25x) - (160 + 10x)
P(x) = 50x - 1.25x^2 - 160 - 10x
P(x) = -1.25x^2 + 40x - 160

To find the maximum profit, you can use the formula x = -b / (2a) if the quadratic equation is in the form ax^2 + bx + c = 0. In this case, a = -1.25, b = 40, and c = -160:

x = -40 / (2 * -1.25)
x = -40 / -2.5
x = 16

Now you have the value of x = 16, which represents the quantity in thousand units that maximizes the profit.

To calculate the maximum profit, substitute x = 16 into the profit function P(x):

P(x) = -1.25(16)^2 + 40(16) - 160
P(x) = -1.25(256) + 640 - 160
P(x) = -320 + 640 - 160
P(x) = 160

Therefore, the maximum profit is 160 thousand dollars (to the nearest thousand dollars), not 160 as you mentioned.

set R(x) = C(x)

x(50-1.25x) = 160 + 10x
50x - 1.25x^2 -160 - 10x = 0
1.25x^2 -40x + 160 = 0
x^2 - 32x + 128 = 0
complete the square
x^2 - 32x = -128
x^2 - 32x + 256 = -128+256
(x-16)^2 = 128
x-16 = ±√128

x = 16± √128 = 4.686 or 27.314 , but 0 < x < 20
so x = 4.686

I will assume your R, C, and P represent
Revenue, Cost and Profit respectively, thus
P(x) = x(50-1/25x) - (160 + 10x)
= 50x - 1.25x^2 - 160 - 10x
P'(x) = 50 - 2.5x - 10
= 0 for a max of P
2.5x = 60
x = 24

max profit = P(24) = ....
I will let you push them buttons.