GATE CSE 2012 | Question: 56

Question

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 $\text₹ 50$ per unit. The number of units to be produced by the firm such that the profit is maximized is

• A. $5$
• B. $10$
• C. $15$
• D. $25$

Comments

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

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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! 

Answer 1

Answer is A.

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

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For example :

5 units =  $CP = 125 \ SP=250 \ \therefore \text{Profit} =125$

10 units =  $CP = 500\ SP=500\therefore \;\text{Profit} =0$ and so forth..

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Therefore, its maximum is at unit $= 5$

Comments

Equation to be maximized:

$f(q) = 50q - 5q^{2}$

First derivative: $f^{\prime}(q) = 50 - 5.2q = 50 - 10q$

At maxima or minima, $f^{\prime}(q) = 0$

$50 - 10q \Rightarrow \boxed{q = 5}$

Now, Second derivative: $f^{\prime\prime}(q) = 0 - 10 = -10$

So, $f^{\prime\prime}(q=5) = -10 = -ve$,

This means $q = 5$ satisfies both conditions for being maxima:

1. $f^{\prime}(q=5) = 0$
2. $f^{\prime\prime}(q=5) = -ve$

Therefore, we have a maxima at $q=5$.

PS: As $f(q)$ is a quadratic equation, we have global maxima at $q = 5$.

Answer 2

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

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

P(x)=50x-5x^{2   ...}here we have to maximize the profit ..differentiate it..

d(P(x))/dx =50-10x =0  ...

x=5     ...option A

Answer 3

Cost Function
          C = 5q²
               │
               │
        Revenue Function
          R = 50q
               │
               ▼
        Profit Function
      π = R − C
      π = 50q − 5q²
               │
               │ Differentiate
               ▼
     dπ/dq = 50 − 10q
               │
               │ Set = 0
               ▼
          50 = 10q
               │
               ▼
            q = 5