3. Problem 2: Suppose a radio manufacturer has the total cost function C (x) = 43x + $ 1850 and the revenue function R (x) = $ 80x. Where "x" is the number of radios manufactured and sold. Find: pli hel

Given that,

Total cost function, C (x) = 43x + $1850

The revenue function R (x) = $80x

To find,

The number of units that must be produced and sold to break even.

Solution,

At break even, cost = revenue

43x + $1850 = $80x

Subtract 80x from both sides.

43x + 1850 -80x = $80x -80x

1850 = 80x-43x

37x = 1850

x = 50

So, the required number of units are 50.

A)  both functions are linear

B) f(x) = 7446.4x + 178713.6 and f(x) = 11000x + 179000

C) House 2 will value $106894.4 more than house 1.

Step-by-step explanation:

A) Value Increase from year 1 to year 2:

House 1: 193,606.40 - 186,160 = 7446.4

House 2: 201,000 - 190,000 = 11000

Value Increase from year 2 to year 3:

House 1: 201,350.66 - 193,606.40 = 7744.26

House 2: 212,000 - 201,000 = 11000  

This means that a constant increament in x variable gives a constant increament in both houses vales. Then, both functions are linear.

B) The slope is the same as the value increment from one year to the next one.

slope (m) of House 1: 7446.4

slope (m) of House 2: 11000

General formula of a line:

f(x) = mx+b

Replacing with a known point:

House 1

186,160 = 7446.4(1) + b

b = 186,160 - 7446.4 = 178713.6

equation: f(x) = 7446.4x + 178713.6

House 2

190,000 = 11000(1) + b

b = 190,000 - 11000 = 179000

equation: f(x) = 11000x + 179000

C) Replacing x = 30 into each equaiton:

Value of House 1 after 30 years

f(x) = 7446.4(30) + 178713.6 = 402105.6

Value of House 2 after 30 years

f(30) = 11000(30) + 179000 = 509000

Then, house 2 will value 509000 - 402105.6 = $106894.4  more than house 1.