How Do You Calculate Marginal and Average Revenue for Television Sales?

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1. Assume that total revenue from the sale of x television sets is given by
R(x)=1000(1-(x/5))^2
a)find the marginal revenue when 400, 500, 600 sets are sold
i found that R'(x)=-4 +(x/125)
is it correct?
therefore R'(400)=$-0.8 R'(500)=$0 R'(600)=$0.8
??

b)find the average revenue form the sales of x sets
ave R(x)=R(x)/x
would it equal (1000/x)(1-(x/500))^2 ??
could it be simplify further?

c) Find the marginal average revenue.
ave R'(x)=R'(x)/x
=(-4/x)+(1/125)

am i correct on all of these??

2. The cost is dollars of manufacturing x items is given by
C=2000x+3500
demand equation is x=√(15000-1.5p)
in terms of the demand x,
a) find an expression for the revenue R

does that mean R(x)=px
R(x)=p√(15000-1.5p)
or u solve for p instead?
 
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1. Assume that total revenue from the sale of x television sets is given by
R(x)=1000(1-(x/5))^2
a)find the marginal revenue when 400, 500, 600 sets are sold
i found that R'(x)=-4 +(x/125)
is it correct?
Nope, applying the chain rule you get (-1/5)*(2)*1000(1-(x/5)) which, upon simplification yields R'(x)=-400+80x
Kevin
 
o..my mistake...the question is actually
R(x)=1000(1-(x/500))^2
then is my answer right now?
 
yep, at least as far as R'(x) is concerned, I don't know any economics so I won't say anything else about the rest of the problem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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