Chapter 2 Homework Key Essay examples

Econ 2994 ch. 2, #1, 3, 7 KEY
1a. Overhead increases by 15 percent.

With the chapter example, the break even was derived as follows:

R = 9.99Q, fixed cost (overhead) = 45,000, and variable cost is 6.99Q
Setting 9.99Q = 45000 + 6.99Q, we found a break even Q of 15000

Here is how to derive the new break even output level:

(FC) Overhead increases by 15% of 45,000 = 6750, and now totals 51,750

(VC) Variable cost is 6.99Q

Profit F = R – C = 9.99Q – (51750 + 3Q)

Break even occurs when F = 0; thus,

F = -51750 + 3Q = 0
3Q = 51750
Q = 17250

Note: as the firm’s costs have increased but its selling price has not, it needs to sell more to achieve the same break even and profit levels.

1b. Costs increase by 1.50 per item.

Again, the C equation is affected, not R. but this time it is the variable cost element of the C equation which changes. We now have:

C = 45,000 + 8.49Q

And require:
F = 9.99Q – (45000 + 8.49Q) = 0
Solving:
Q = 45000/1.50 = 30000 as the new break-even level of output.

1c. Selling price increases to 11.99

Now R changes while C does not, giving:
F = 11.99Q – (45000 + 6.99Q) = 0

Solving:
Q = 45000/5 = 9000 as the new break-even level of output.

3a. We have a linear demand equation Qd = 1000 – 5P
To graph a linear equation, we need two sets of co-ordinates. Our author (unlike economists) puts Qd on vertical Y axis and P on the horizontal X axis.
From the problem, price can vary from 0 to 200.
Setting P = 0 and solving, then Qd = 1000 – 5(0) = 1000
Setting P = 200, Qd = 1000 – 5(200) = 0
We could have set Qd = 0 to solve for P as well;
0 = 1000 – 5P
5P = 1000
P = 200
So, our two sets of coordinates are P = 0, Qd = 1000 and Qd = 0, P = 200
Connecting these two coordinates graphs our equation.
3b. The downward or negative slope of this curve tells us something that is commonsense: quantity demanded (one of our variables) is negatively (inversely) related to the other: price.
3c. If (total) revenue is price per item times quantity sold, then
R = Qd (P)
Since Qd = 1000 – 5P, we have
R = Qd (P) = (1000 – 5P)(P) = 1000P – 5P2
3d. To graph this ‘quadratic’ equation requires several sets of points.
P Q
0 0
20 18000
40 32000
60 42000
80 48000
100 50000
120

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