Question

A food company produces $ x $ (hundred) quality A and y (hundred) quality B items per day, where $ y(5 - x) = 10(4 - x),\,\,0 \leqslant x \leqslant 4 $ . If the profit on each quality A item is twice the profit on quality B item, then the most profitable number of quality A item per day to manufacture are
( $ \sqrt 5 = 2.24) $

Answer 1

Hint: Derive equation for both the items and use profit condition.
 We are going to use the given equation for getting equation for the y items and obtaining the equation, we are going to use the condition given about the profits of items and then we get an equation for the total profits which we will differentiate it with respect to time and then we will get the required value of quality A items to be manufactured.

Complete step by step solution:
We are given that
 $ y(5 - x) = 10(4 - x) $
We are going to get the equation such that it equates to y.
We will get
 $ y = \dfrac{{40 - 10x}}{{\left( {5 - x} \right)}} $
So, number of products A= $ x $ and
The number of products B $ y = \dfrac{{40 - 10x}}{{\left( {5 - x} \right)}} $
We are given in the question that the profit on each quality A item is twice the profit on quality B item, so from that we get an equation for total profit
 $ Total = \,2\left( x \right) + \dfrac{{40 - 10x}}{{\left( {5 - x} \right)}} $
We are going to simplify this equation,
 $
  Total = \dfrac{{2x\left( {5 - x} \right) + \left( {40 - 10x} \right)}}{{\left( {5 - x} \right)}} \\
   = \dfrac{{10x - 2{x^2} + 40 - 10x}}{{5 - x}} \\
   = \dfrac{{ - 2{x^2} + 40}}{{5 - x}} \;
  $
Since, we have to find the number of products A per day, we are going to differentiate with respect to $ x $ , which gives us

 $ \dfrac{{dp}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{ - 2{x^2} + 40}}{{5 - x}}} \right) $
We are going to use the $ \dfrac{u}{v} $ derivation method and solve it, which gives us
 $ \dfrac{{uv' - vu'}}{{{v^2}}} $
So, we get after simplification
 $
   = \dfrac{{\left( {5 - x} \right)x( - 4x) - (40 - 2{x^2})x( - 1)}}{{{{(5 - x)}^2}}} \\
   = - 20x + 4{x^2} + 40 - 2{x^2} \;
 $
We are going to equate this to zero, to solve the equation
 $
   - 20x + 4{x^2} + 40 - 2{x^2} = 0 \\
  2{x^2} - 20x + 40 = 0 \\
  {x^2} - 10x + 20 = 0 \;
  $
We are going to find the root with the help of the formula
 $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
We will substitute in the formula and get the roots
 $
   = \dfrac{{10 \pm \sqrt {{{10}^2} - 4(1)(20)} }}{{2(1)}} \\
   = 5 \pm \sqrt 5 \;
 $
We know the value of $ \sqrt 5 $ , we substitute there
 $
   = 5 \pm (2.24) \;
 $
We get two values which are \[2.76\] and \[7.24\] but we know that $ 0 \leqslant x \leqslant 4 $ ,
So, we consider
 $ x = 2.76 $
We know that $ 100x = 100 \times 2.76 = 276 $ products need to be manufactured.
So, the correct answer is “276”.

Note: We have to read the question carefully as we need to substitute the x in $100x$ to get the number of products to be manufactured daily to be in a profitable manner and we should form the equations carefully.