Find the largest possible revenue from the demand equation "Q= -2p + 1000

1 answer:

3 0

Revenue=quantity x price= (-2p+1000) (p)= -2p^2+1000p
The maximum revenue will occur when the first derivative is zero so when 2(-2p)+1000=0;p=250
Which generates 125,000 in revenue
Try prices of 245 and 255 and you will see they both are less than 250 thereby proving the max revenue is 250

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Integration: How would I get to this answer?

Scrat [10]

Given that $y$ attains a maximum at $x=1$ , it follows that $y'=0$ at that same point. So integrating once gives

$\displaystyle\int\frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx=\int-8x\,\mathrm dx$
$\dfrac{\mathrm dy}{\mathrm dx}=-4x^2+C_1$
$\implies -4(1)^2+C_1=0\implies C_1=4$

and so the first derivative is $y'=-4x^2+4$ .

Integrating again, you get

$\displaystyle\int\frac{\mathrm dy}{\mathrm dx}\,\mathrm dx=\int(-4x^2+4)\,\mathrm dx$
$y=-\dfrac43x^3+4x+C_2$

You know that this curve passes through the point (2, -1), which means when $x=2$ , you have $y=-1$ :

$-1=-\dfrac43(2)^3+4(2)+C_2$
$\implies C_2=\dfrac53$

and so

$y=-\dfrac43x^3+4x+\dfrac53$