# Thread: Consumers' Surplus for Calculus

1. ## Consumers' Surplus for Calculus

The demand function for a certain make of replacement cartridges for a water purifier is given by:
p=-0.01x^2-0.1x+6
where p is the unit price in dollars and x is the quantity demanded each week, measured in units of a thousand.
Determine the consumers' surplus if the market price is set at $4/cartridge. I know the answer is$11,667. How did they get this answer? Particularly, how would I find x for the bounds when I integrate? Please show steps in a thorough, understandable manner.
(The problem involves integration.)
Thank you.

(If anyone asks for a supply function or any second equation -- there is none. This is how the question has been exactly written in the book.)

2. ## Re: Consumers' Surplus for Calculus

Problem makes no sense. Why is the problem asking for a surplus (units: cartridges) when the answer is a dollar amount?

5. ## Re: Consumers' Surplus for Calculus

But that's not the answer in the book. *Smiles* So, are you saying that the book's answer is wrong?

And so, if I set 4 to the equation, it would be:
4= -0.01x^2-0.1x+6
Subtract 4 to get to the other side, which would make: -.0.01x^2-0.1x+2
Is that what you mean?

Thank you.

6. ## Re: Consumers' Surplus for Calculus

Sorry, I might've mis-understood! I didn't know the difference between consumers' surplus and a surplus (they're two completely different things).

According to Wikipedia, a consumers' surplus is "the monetary gain obtained by consumers because they are able to purchase a product for a price that is less than that they would be willing to pay." Using this definition, we find the highest price that consumers are willing to pay by maximizing $p$. To do this, we graph p and see that the highest p that produces a non-negative demand is p = 6.

The consumers' surplus is equal to

$S = \int_{4}^{6} x(p)\, dp$ where x(p) is the demand as a function of price. To find x(p) we must solve for x in the equation:

$p = -0.01x^2 - 0.1x + 6 \Rightarrow 0.01x + 0.1x + (p-6) = 0$

$x = \frac{-0.1 + \sqrt{0.01 - 4(0.01)(p-6)}}{0.02} = -5 + 5\sqrt{25 - 4p}$ (taking positive root).

However x is in thousands so we should multiply this expression by 1000 to get $1000(-5 + 5\sqrt{25 - 4p})$. Integrating from p = 4 to p = 6,

$S = \int_{4}^{6} 1000(-5 + 5\sqrt{25-4p}) \, dp = \frac{35000}{3} = 11667$.

Aha! Previously I didn't know the difference between "consumers' surplus" and "surplus." Sorry for the prior confusion.

7. ## Re: Consumers' Surplus for Calculus

Aha! Previously I didn't know the difference between "consumers' surplus" and "surplus." Sorry for the prior confusion.
No need for an apology ... it is incumbent on the original poster to clearly interpret all contextual terms used in posting a problem.

8. ## Re: Consumers' Surplus for Calculus

I'm sorry, but I'm still confused. I've been going over the quadratic equation many times over and trying to figure out how you received -5+5(sq/25-4p)

I've never solved a quadratic equation with an actual letter in it, so I was confused as to how to go about it. I thought to substitute 1 for a p so it would be (1-6), which within the square root of 0.1^2-4(0.01)(p-6) would equal square root 0.21
Many thanks if you would explain this part to me.

And don't worry about confusing the "surplus" aspect, we all mess up sometimes.
Good day.

9. ## Re: Consumers' Surplus for Calculus

$p=-0.01x^2-0.1x+6$

$0.01x^2 + 0.1x + p - 6 = 0$

$a = 0.01$

$b = 0.1$

$c = p - 6$

$x = \frac{-0.1 + \sqrt{(0.1)^2 - 4(0.01)(p-6)}}{2(0.01)}$

$x = \frac{-0.1 + \sqrt{0.01 - 4(0.01)(p-6)}}{0.02}$

$x = \frac{-0.1 + \sqrt{0.01[1 - 4(p-6)]}}{0.02}$

$x = \frac{-0.1 + 0.1\sqrt{25 - 4p}}{0.02}$

$x = \frac{-0.1[1 - \sqrt{25 - 4p}]}{0.02}$

$x = -5 \left(1 - \sqrt{25-4p} \right)$

$x = 5 \left(\sqrt{25-4p} - 1 \right)$

10. ## Re: Consumers' Surplus for Calculus

Originally Posted by AntoninaFinn
I'm sorry, but I'm still confused. I've been going over the quadratic equation many times over and trying to figure out how you received -5+5(sq/25-4p)
Ah, good algebra skills are essential for a calculus course. Just use the quadratic formula and substitute a = 0.01, b = 0.1, c = p-6 as skeeter did.