Question

I need help with part 2 of this problem. I have already solved part one... unless it is wrong. Part 1 is a theoretical question and asked me to make up the numbers. I basically did columns A and B like so (listed below).

x	f(x)
1	10
2	9
3	8
4	7
5	6
6	5
7	4
8	3
9	2
10	1

Part 1:

a. [4] …derive the product’s linear (1st order) trend line equation C = f(x).
[Answer]:	y = (-x) + 11
b. [2] …identify the model’s coefficient of determination—its “[(R^2)]”.
[Answer]:	R^2 = 1

Please help with part 2, a 6 part question:

1. Write the product’s linear Revenue “R” model, assuming this
product’s current unit price “p” remains constant over time at $850.
2. Write the product’s linear Profit “P” model, based on the Cost and Revenue models you derived above.
3. Write the numerical value for the slope “m” of the “C”, “R” and “P”
models. Interpret what each value indicates about the magnitude
…and…direction of change in the dependent variable expected to be
associated with a 1-unit change in the independent variable.
4. Write the numerical value for the vertical intercept of the “C”, “R”
and “P” models and interpret each value.
5. Forecast “C”, “R” and “P” @ “x” = 150.
6. Calculate how many units “x” of the product must be produced and
sold in order to generate a Profit of $150,868.

Answer 1

2) As its given that the product’s current unit price “p” remains constant over time at $850.

So $R(x)=850x$

The profit function is given by

PC) = R(2) - C() = 850.0 - (-1 +11) = 851.0 11

3) Slope of a linear model mx+c, is the coefficient of x.

Slope of is , which means that the cost decreases as the number of product sold increases

Slope of is 850, So the revenue increases 850 times as the number of products sold increases

Slope of is 851, So the Profit increases 851 times as the number of products sold increases.

4)

Vertical intercept of is 11, the basic cost is 11

Vertical intercept of is 0, So if no product is sold there is no revenue.

Vertical intercept of is , So if no product is sold there is a loss of 11

5) For x=150, we get

$C(150)=(-150)+11=-139 \\~~~~~ R(150)=850\times 150=127500 \\~~~~~ P(150)=851 \times 150-11=127,639$

6) P= 150,868  

Implies

$150,868=851 x-11 \\ \Rightarrow 851x=150,879 \\ \Rightarrow x\approx 177.296$

So approximately 178 product must be produced and sold in order to generate a Profit of $150,868