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A company is planning to manufacture and market a new headphone set. After conducting extensive market surveys, the research department provides the following estimates:

Marginal costs function: $c'(x)=20+0.2x$ where $x$ is the quantity sold. Fixed cost:500

Demand function is linear :At the price of $78$, 30 units were sold. At the price of $58$,$50$ units were sold.

Find the following a) Assume that the relationship between price $p$ and demand $x$ is linear . Express $p$ as a function of $x$.

b)Find the total cost function

c)Find the profit function

d)The level of production in which profit is maximized.

e) The maximum profit, price per unit, total revenue and total cost when profit is maximized.

f) Find the change in total profit when production level increases from 50 units to 60 units. Interpret your results.

My attempt,

a) $p=mx+b$ $m=\frac{50-78}{58-30}=-1$ $p=-x+b$ $50=-58+b$

So, $p=-x+108$

b)Total cost function$=\int 20+0.2xdx$

$20x+\frac{0.2x^2}{2}+500=0.1x^2+20x+500$

c)$P(x)=xp(x)-c(x)=x(-x+108)-(0.1x^2+20x+500)$

$P(x)=-1.1x^2+88x-500$

d)$\frac{dP}{dx}=0$

$\frac{d}{dx}(-1.1x^2+88x-500)=0$

$88-2.2x=0$

$x=40$

Am I correct for my attempts? How to proceed for e) ? Can anyone give me some tips for me? Thanks

$R(x)=xp(x)=(-x+108)x=-x^2 +108x$

To find the level of production in which revenue is maximised.

$\frac{d}{dx}(-x^2+108x)=0$

$-2x+54=0$

$x=27$

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Your (a) is not correct. Although the answer is correct. $p$ is dependent variable, $x$ is independent variable. The slope $m$ should be $$\frac{58-78}{50-30}$$ To find $b$, you plugged the wrong numbers: $58=-50+b$

The maximum profit is $P(40)$. You can find it by plugging $40$ into the profit function.

Price per unit: plug $40$ into $p(x)$.

Total cost: plug $40$ into $c(x)$.

Total revenue: $40*p(40)$.

Part (f): $P(60)-P(50)$.

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  • $\begingroup$ If I found the level of production in which revenue is maximised which is $27$. How to find total revenue? $\endgroup$ – Mathxx Jun 5 '15 at 9:27
  • $\begingroup$ Can I know how to find the break-even of the production level also? I take $R(x)=C(x)$ I got $1.1x^2-88x+500=0$. Then? @KittyL $\endgroup$ – Mathxx Jun 5 '15 at 9:35
  • $\begingroup$ @Mathxx: How did you get $27$ and why do you want to maximize revenue? For the break-even, you can solve the quadratic equation by quadratic formula. $\endgroup$ – KittyL Jun 5 '15 at 9:56
  • $\begingroup$ I've edited the post. How to find the total revenue then ? I got two values for even break. Which one should I choose? $\endgroup$ – Mathxx Jun 5 '15 at 10:09
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    $\begingroup$ @Mathxx: Check your derivative, the production that maximizes revenue should be $54$. But the question didn't ask you find that. Or is it the next question? To find the total revenue, when the profit is maximized, just plug $40$ into the revenue function. There could be two break-even points. $\endgroup$ – KittyL Jun 5 '15 at 10:56

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