A manufacturer can sell x items at a price of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $R(x)$ be the revenue function and $C(x)$ be the cost function. Revenue $R(x) = x 	imes \left(5 - \frac{x}{100}ight) = 5x - \frac{x^2}{100}

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(B) Let $R(x)$ be the revenue function and $C(x)$ be the cost function. Revenue $R(x) = x 	imes \left(5 - \frac{x}{100}ight) = 5x - \frac{x^2}{100}$. Profit function $P(x) = R(x) - C(x) = \left(5x - \frac{x^2}{100}ight) - \left(\frac{x}{5} + 500ight)$. $P(x) = 5x - \frac{x^2}{100} - \frac{x}{5} - 500 = 4.8x - \frac{x^2}{100} - 500$. To find the maximum profit,we find the derivative $P'(x)$ and set it to $0$. $P'(x) = 4.8 - \frac{2x}{100} = 4.8 - \frac{x}{50}$. Setting $P'(x) = 0$,we get $4.8 = \frac{x}{50}$,which implies $x = 4.8 	imes 50 = 240$. To verify,we check the second derivative $P''(x) = -\frac{1}{50}$. Since $P''(x) < 0$,the profit is maximum at $x = 240$.

$A$ manufacturer can sell $x$ items at a price of rupees $\left(5 - \frac{x}{100}\right)$ each. The cost price of $x$ items is Rs. $\left(\frac{x}{5} + 500\right)$. The number of items that the manufacturer should sell to earn the maximum profit is

Similar Questions

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$.

Find the local maximum and local minimum values for the function $f(x) = \sin x - \cos x$ where $0 < x < 2\pi$.

The maximum value of $\frac{\log _{e} x}{x}$,if $x>0$ is

If $f(x) = \frac{\sin(x + a)}{\sin(x + b)}$,$a \neq b$,then $f$ is......

If $x$ and $y$ are two positive integers such that $x + 2y = 10$ and $x^2 y^3$ is maximum,then $x^2 + 2y^3 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module