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The cost function for a product in a firm is given by $5q^{2}$, where $q$ is the amount of production. The firm can sell the product at a market price of $₹ 50$ per unit. The number of units to be produced by the firm such that the profit is maximized is

1. $5$
2. $10$
3. $15$
4. $25$
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cost function is for 1 product and selling price is also for 1 product so why to multiply q with 50??
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gari

here question is for a single product and number of unit of this product....q is the amount of unit of production

and sell price is 50 for given product and per unit...

Profit(x):- Selling price(50*x) -production cost( 5x2 )

P(x)=50x-5x2

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@hs_yadav .. cost price for a single product is 5x2 and selling price for a single product is 50 ...

so profit = 50-5x2. (per product)...(I am saying this according to the language in the question)...?

where am i wrong?

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The firm can sell the product at a market price of ₹ 50 per unit

therefor selling price ... 50x   not 50

+1

Profit = S.P - C.P
$CP = 5Q^2$
$S.P = 50Q$
If company produces only one item then
$C.P=5*1^2=5$
$S.P =50*1=50$, $profit=50-5=45 Rs.$
if Company Produces 2 items then Q=2
$C.P = 20, SP=100, Profit=80 Rs$
if Company Produces 5 items then Q=5
$C.P = 125, SP=250, Profit=125 Rs$
if Company Produces 6 items then Q=6
$C.P = 180, SP=300, Profit=120 Rs$

Note that when Q=5, profit was 125 Rs and when Q=6 then profit is 120 Rs.
Hence, 5 is the correct choice!

The equation for profit is Profit=$SP-CP,$

here $SP=Q\times 50$  and $CP=5Q^{2}$.

So when a function attains its maximum value its first order differentiation is zero.

Hence $50-5\times 2\times Q=0 \therefore Q=5.$

For example :

5 units =  $CP = 125 \ SP=250 \ \therefore Profit=125$

10 units =  $CP = 500\ SP=500\therefore Profit=0$ and so forth..

Therefore its maximum at unit $= 5$

edited

let x is the number of unit at which profit is maximized......

Profit(x):- Selling price(50*x) -production cost( 5x2 )

P(x)=50x-5x2   ...here we have to maximize the profit ..differentiate it..

d(P(x))/dx =50-10x =0  ...

x=5     ...option A

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