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Find the Consumer Surplus, given the demand and supply equations $$ D(x)=\frac{405}{\sqrt{x}} $$ $$ S(x)=5\sqrt{x} $$ The equilibrium point is $(81,45)$.

I know the formula for consumer surplus, but I am stuck on finding the integral of $405/\sqrt{x}$.

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For context, I grabbed this picture from Wikipedia

surplus

The red area is the integral of $D(x) - 45$ from $0$ to $81$. Namely, $$ \int_0^{81} \left(\frac{405}{\sqrt{x}}-45\right)\,dx $$ To integrate, write $405/\sqrt{x}$ as $x^{-1/2}$ and use the formula $\int x^a = x^{a+1}/(a+1)$.

To check the answer, you can use Wolfram Alpha: it's $3645$.

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Point of Equilibrium for Supply and Demand

$$ D(q_e)=S(q_e) $$ $$ \frac{405}{\sqrt{q_e}}=5\sqrt{q_e} $$ $$ 405=5|q_e|$$ $$ |q_e|=\frac{405}{5}=81$$ The next step is to find the market price, which is $$ P_{mkt}(q_e)=D(q_e)=S(q_e)=5\sqrt{81}=5\cdot 9=45 $$ Therefore, the point of equilibrium is $$ (q_e, P_{mkt}(q_e))=(81, 45)$$ Consumer Surplus

By definition the consumer surplus $CS$, is the area between the demand curve and the market price from $0$ to the quantity at the point of equilibrium. Mathematically this is expressed as $$ CS= \int_{0}^{q_e} D(q) - P_{mkt}(q_e)\ dq $$ So in this case we have, $$ CS= \int_{0}^{81} \frac{405}{\sqrt{q}} - 45\ dq $$ $$= 405\int_{0}^{81} q^{-\frac{1}{2}}dq- 45\int_0^{81} dq $$ $$= 405\left[\frac{q^{\frac{1}{2}}}{\frac12}\right]_{0}^{81} - 45(81-0)$$ $$= 2(405)(\sqrt{81}-0)- 45(81-0)$$ $$= 2(405)(9)- 45(81) $$ $$=7290-3645=3645 $$

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  • $\begingroup$ I have that part done, but im stuck on finding the consumer surplus. Right now I am trying to find (405/sqrt(x))dx $\endgroup$
    – Ben
    Nov 24 '14 at 20:15
  • $\begingroup$ You know what I'm saying? $\endgroup$
    – Ben
    Nov 24 '14 at 20:26
  • $\begingroup$ @Rafflesiaarnoldii I was agreeing that what he answered was correct $\endgroup$
    – Ben
    Nov 24 '14 at 20:48
  • $\begingroup$ @Rafflesiaarnoldii, that's why it's called a hint. By the way, his equilibrium point wasn't posted when I posted it. $\endgroup$
    – k170
    Nov 24 '14 at 20:50
  • $\begingroup$ After that step would it be: [405(81)^(-1/2)-405(0)^(-1/2)]-3645? $\endgroup$
    – Ben
    Nov 24 '14 at 20:56

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