Question

The total cost of producing $'x'$radio sets per day is Rs. $\left( \dfrac{{{x}^{2}}}{4}+35x+25 \right)$and price per set at which they may be sold is Rs. $\left( 50-\dfrac{x}{2} \right)$. Find the daily output to maximise the total profit.

Answer 1

We solve this problem first by calculating the profit. If C.P is the cost price and S.P is the selling price of $'x'$radio sets per day, then the profit is calculated by using the formula
$P=S.P-C.P$
Here, as P is the function of $'x'$, to find the output of maximum profit we solve the value of $'x'$by making the differentiation of P to zero that is
$\Rightarrow \dfrac{d}{dx}\left( P \right)=0$
The value of $'x'$from the above equation will give the required answer.

Complete step-by-step answer:
We are given that the cost price of $'x'$radio sets per day is
$\Rightarrow C.P=\left( \dfrac{{{x}^{2}}}{4}+35x+25 \right)$
We are given that selling price of one radio set is $\left( 50-\dfrac{x}{2} \right)$
Now, let us calculate the total selling price of $'x'$radio sets per day as follows

& \Rightarrow S.P=x\left( 50-\dfrac{x}{2} \right) \\\ & \Rightarrow S.P=50x-\dfrac{{{x}^{2}}}{2} \\\ \end{aligned}$$ We know that the profit is calculated by using the formula $$P=S.P-C.P$$. By using the above formula we get the profit as $$\Rightarrow P=\left( 50x-\dfrac{{{x}^{2}}}{2} \right)-\left( \dfrac{{{x}^{2}}}{4}+35x+25 \right)$$ By dividing the terms of same power we get $$\begin{aligned} & \Rightarrow P=\left( -\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{2}}}{4} \right)+\left( 50x-35x \right)-25 \\\ & \Rightarrow P=-\dfrac{3{{x}^{2}}}{4}+15x-25 \\\ \end{aligned}$$ Now, let us find the value of $$'x'$$for which the profit P is maximum. We know that, to find the value of $$'x'$$for which the function is the maximum is calculated by solving the equation $$\Rightarrow \dfrac{d}{dx}\left( P \right)=0$$ By substituting the function P in above equation we get $$\Rightarrow \dfrac{d}{dx}\left( -\dfrac{3{{x}^{2}}}{4}+15x-25 \right)=0$$ Now, by dividing the terms we get $$\begin{aligned} & \Rightarrow -\dfrac{3}{4}\left( \dfrac{d}{dx}\left( {{x}^{2}} \right) \right)+15\dfrac{d}{dx}\left( x \right)-25\dfrac{d}{dx}\left( 1 \right)=0 \\\ & \Rightarrow -\dfrac{3x}{2}+15=0 \\\ & \Rightarrow x=\dfrac{15\times 2}{3}=10 \\\ \end{aligned}$$ Therefore, we can say that the profit is the maximum when there is a daily output of 10 radio sets. **Note:** Students will make mistakes in selling the price of the radio sets. We are given the selling price of one radio set as $$\left( 50-\dfrac{x}{2} \right)$$. But students will consider this price as the selling price of $$'x'$$radio sets which will be wrong. They consider $$S.P=\left( 50-\dfrac{x}{2} \right)$$which leads to wrong answers. So, reading the question and considering the values is important.