Reading 12 Demand and Supply Analysis (Marginal revenue)


Can someone help me with the below question?

Q: Company A is the only producer of widgets. At a price of $10, the quantity of widgets demanded is 200, while at a price of $8, the quantity demanded is 300. What is the marginal revenue Company A receives when production is at 200 widgets?

A: $6
Hint was given: Use the formula: marginal revenue = price - (quantity * slope)

Assuming a linear relationship between Price and Quantity:
Price = a + b \times Quantity

Slope, b = \frac{\Delta Price}{\Delta Quantity} = \frac{\$8-\$10}{300-200} = -0.02

Price = a - 0.02 \times Quantity

Substituting Price = $10 and Quantity = 200, we get:

10 = a - 0.02 \times 200

a = 10 + 4 = 14


Price = 14 - 0.02 \times Quantity

or just: P = 14 - 0.02Q

Total ~Revenue = Price \times Quantity

Total ~Revenue = (14 - 0.02Q) \times Q

Total ~Revenue = 14Q - 0.02 Q^2

Marginal ~Revenue, MR = \frac{d}{dQ}(TR) = 14(1) - 0.02(2Q) = 14 - 0.04Q

At Q = 200, then MR = 14 - 0.04(200) = \$6

Thanks millions for such a wonderful explanation! If I had done it this way ($8-(100@ ($8-$10)/300-200))) as per the formula, would it be incorrect? Given that we will have only 90 secs to solve the question.

Also, I got confused at Marginal Revenue step: 14(1)-0.02(2Q). Where did 1 and 2 come from in this equation? I guess, I am lacking math skills here. Could you please help me with this?

Many many thanks!

This will work out to be:

8 - 100 \times (\frac{-2}{100}) = 8 + 2 = \$ 10

which is not the answer you are looking for. And the question requires the marginal revenue at quantity = 200, so that’s not correct too.

The right formula should be:

Marginal ~revenue = Price + Quantity \times Slope

Then when Price = $10, and Quantity = 200, we have

Marginal ~revenue = 10 + 200 \times (\frac{8-10}{300-200})

Marginal ~revenue = 10 + 200 \times (-0.02) = \$6

Differentiation from calculus. General formula for differentiating terms with power:

\frac {d}{dQ} (Q^n) = nQ^{n-1}


\frac {d}{dQ} (Q^2) = 2Q^{2-1} = 2Q

\frac {d}{dQ} (Q^1) = 1Q^{1-1} = 1Q^0 = 1

Wow, thanks so much for explaining it so well! I definitely lack calculus skills.

You don’t need to know calculus for this exam.

1 Like

thanks! But, I would still need skills of solving equations. Any thoughts, what basis course I could do online to get upto speed?

The difficulty with this formula is how they define “slope”. As fino (correctly) calculated it,

slope = \frac{\Delta P}{\Delta Q} = \frac{$8 - $10}{300\ widgets - 200\ widgets} = -$0.02/widget

However, that doesn’t work in the formula that they gave you in their hint:

MR = P - (Q Ă— slope)

As fino pointed out, the correct formula (using slope calculated correctly) is:

MR = P + (Q Ă— slope)

In my 2019 copy of the Level I curriculum, this formula is shown in the reading Topics in Demand and Supply Analysis, §3.2.2.

It appears that the author of your question wanted you to use this formula for the slope:

slope = \left\lvert\frac{\Delta P}{\Delta Q}\right\lvert = \left\lvert\frac{$8 - $10}{300\ widgets - 200\ widgets}\right\lvert = $0.02/widget

In that case, their formula is correct.

Note that, in either case, you’re simply using the formula given; you don’t need calculus.

Where did you get this question, if I may ask?

Many thanks for explaining. To be honest, when I couldn’t understand Fino’s initial equation explaination that’s when I thought I would need more mathematical skills than I have.

I got this question from online,

Another question: I am registered to write CFA level-1 in Dec 2020. I was working as an associate equity analyst specializing in mining companies in Vancouver (BC) and lost my job recently. I want to change the industry and I am thinking of doing CPA on the side while I am doing CFA. Would you have any thoughts?

Well, Analystnote got the formula wrong.

CFA Institute won’t.

1 Like