Question

Total cost of producing x pockets radio sets per day is Rs $\left( \dfrac{1}{4}{{x}^{2}}+35x+25 \right)$ and the price per set at which they may sold is Rs $\left( 50-\dfrac{x}{2} \right)$ to obtain maximum profit the daily profit output should be _________ radius set.
$\begin{align}
  & a)10 \\
 & b)5 \\
 & c)15 \\
 & d)20 \\
\end{align}$

Answer 1

Hint: We are given total cost of production of x radios as Rs $\left( \dfrac{1}{4}{{x}^{2}}+35x+25 \right)$and price per set of Rs. $\left( 50-\dfrac{x}{2} \right)$, we are asked to find the maximum profit, we will learn how profit is obtained from revenue and cost, we will find critical point, we will use the second derivative test to check the maximum or minimum, we will use that $d({{x}^{n}})=n{{x}^{n-1}}$ to get the value of x which gives maximum profit.

Complete step by step solution:
We are given that total cost of producing item daily is Rs. $\left( \dfrac{1}{4}{{x}^{2}}+35x+25 \right)$ and selling price per unit is Rs. $\left( 50-\dfrac{x}{2} \right)$
We have to find the number of output daily deliveries to get maximum profit. To do so we first learn about revenue cost function and cost function is the function which give us the knowledge about the cost of the production of item, denoted by $(x)$
$C(x)$ is given to us as,
$C(x)=\left( \dfrac{1}{4}{{x}^{2}}+35x+25 \right)$
Reverse function will tell us about the amount earned by selling an item, it is denoted
$R(x)$
As selling price of an item is $\left( 50-\dfrac{x}{2} \right)$ and x price are sold daily so total selling would be of cost $x\times \left( 50-\dfrac{x}{2} \right)$
Hence total revenue is $x\left( 50-\dfrac{x}{2} \right)$
So, we get
$R(x)=x\left( 50-\dfrac{x}{2} \right)$
Now profit is the difference of revenue and the cost. So profit function is denoted as $P(x)$
\[P(x)=R(x)-C(x)\]
Using above value, we get
$P(x)=x\left( 50-\dfrac{x}{2} \right)-\left( \dfrac{{{x}^{2}}}{4}+35x+25 \right)$
Simplifying we get
\[\Rightarrow P(x)=50x-{{\dfrac{x}{2}}^{2}}-{{\dfrac{x}{4}}^{2}}-35x-25\]
By adding like terms, we get,
\[\Rightarrow P(x)=\dfrac{-3}{4}{{x}^{2}}+15x-25\]
Now to find the point x, at which profit is maximum, we will find the central point for which we first differentiate $P(x)$ and then put it as zero.
So $P(x)=\dfrac{-3}{4}(2x)+15-0$
As $d({{x}^{2}})=2x\,\,\,\,\,and\,\,\,\,d(x)=1$
\[\Rightarrow P(x)=\dfrac{-3}{2}x+15\]
Now consider \[P'(x)=0\], and solve for x.
\[\Rightarrow \dfrac{-3}{2}x+15=0\]
\[\Rightarrow x=\dfrac{15\times 2}{3}=10\]
Now we apply second derivative test to check whether it is point of maximum or not we denote it by:
\[P''(x)\]
We get \[P''(x)=\dfrac{-3}{2}\]
Which is less than \[0\]
So, \[x=10\] is the maximum.

So, the correct answer is “Option a”.

Note: Remember that profit is not obtained by subtracting cost of \[10-\dfrac{x}{2}\] or \[50-\dfrac{x}{2}\] is not revenue, it is just the selling price of one item, but of profit we need selling cost of item hence we multiply it by x to get \[R(x)\], also remember in second derivative is positive critical point in point of minimum and if second derivative is negative the critical point is point of maximum.