Question

A company has for x items produced the total cost C and the total revenue R

given by equations C = 100 + 0.015x and R = 3x. Find how many items be

produced to maximize the profit ? What is this profit ?

(d) If the demand function for a product is given by p = 4 - 5x, for what value

of x does the demand curve have unit elasticity ?​

Answer 1

Answer:

ple helped

Given P=Rs.12000N=2R=6%

S.I.=

100

PNR

SI =

100

12000×6×2

=1440

A=SI+P=1440+12000= Rs. 13440

Compound interest:

A=P(1+

100

r

)

n

=12000(1+

100

6

)

(2)

=12000(1.1236)

=Rs.13483.2

CI=A−P=13483.20−12000

=Rs.1483.20

Extra interest =13483.20−1344

Step-by-step explanation:

Answer 2

Answer:

Maximum profit is $50.$

Step-by-step explanation:

Given,

Cost function $C(x)=100+0.015x^2$

Revenue function $R=3x$

Let,  $P(x)=$ Profit function $= R(x)-C(x)=3x-100-0.015x^2$

Therefore,  $\frac{d}{dx} P(x)=3-0.03x$  and  $\frac{d^2}{dx^2} P(x)=-0.03$

For maxima/minima,  $\frac{d}{dx}P(x)=0$ ⇒ $3-0.03x=0$ ⇒ $x=\frac{3}{0.03}=100$

Therefore, at $x=100;$  $\frac{d^2}{dx^2} P(x)=-0.03 < 0$

Thus, profit $P$ is maximum when $x=100$

and hence required no. of items to get maximum profit be $10.$

Also maximum profit $=P(100)=300-100-0.015*10000=300-100-150=50$