Question

What is the maximum value of \[Z\] where, \[Z = 4x + 2y\] subject to constraints \[4x + 2y \ge 46\], \[x + 3y \le 24\] and \[x,y \ge 0\]?
A. 46
B. 96
C. 52
D. None of these

Answer 1

Hint: First we will solve the inequalities in as an equation. Then draw a diagram using the inequalities. The common region of inequality is known as the feasible region. Then we will find the corner points of the feasible region and substitutes the point in \[Z = 4x + 2y\]. The maximum value of \[Z\] is the required solution.

Formula used:
The intercepts form of a line is \[\dfrac{x}{a} + \dfrac{y}{b} = 1\].

Complete step by step solution
Given that,
\[Z = 4x + 2y\]
subject to constraints
\[4x + 2y \ge 46\]
\[x + 3y \le 24\]
\[x,y \ge 0\]
Now rewrite the inequalities as an equation
\[4x + 2y = 46\] ….(i)
\[x + 3y = 24\] ……(ii)
Rewrite the equation intercepts form
For equation (i)
\[\dfrac{{4x}}{{46}} + \dfrac{{2y}}{{46}} = 1\]
\[ \Rightarrow \dfrac{x}{{11.5}} + \dfrac{y}{{23}} = 1\]
The x-intercept is 11.5 and y-intercept is 23.
For equation (ii)
\[\dfrac{x}{{24}} + \dfrac{{3y}}{{24}} = 1\]
\[ \Rightarrow \dfrac{x}{{24}} + \dfrac{y}{8} = 1\]
The x-intercept is 24 and y-intercept is 8.
Now finding the intersection point equation (i) and (ii)
Multiply 4 with equation (ii) and subtract it from equation (i)
\[4x + 2y = 46\]
\[4x + 12y = 96\]
\[\overline {\,\,\,\, - 10y = - 50} \]
\[ \Rightarrow y = 5\]
Putting \[y = 5\] in the equation (ii)
\[x + 3 \cdot 5 = 24\]
\[ \Rightarrow x = 9\]
The intersection point is \[\left( {9,5} \right)\].
Putting \[\left( {0,0} \right)\] in \[4x + 2y \ge 46\].
\[0 + 0 \ge 46\] which incorrect.
So the region of the \[4x + 2y \ge 46\] is side in which \[\left( {0,0} \right)\] does not lie.
Putting \[\left( {0,0} \right)\] in \[x + 3y \le 24\].
\[0 + 0 \le 24\] which correct.
So the region of the \[x + 3y \le 24\] is side in which \[\left( {0,0} \right)\] lies.
Given that x>0 and y > 0.
So the feasible region must lie in the first quadrant.
Now we will draw a diagram of inequality.

Image: Feasible region

Hence the corner points of the feasible region are \[\left( {9,5} \right)\], \[\left( {24,0} \right)\] and \[\left( {11.5,0} \right)\]
Now putting \[x = 9\] and \[y = 5\] in the equation \[Z = 4x + 2y\]
\[Z = 4 \cdot 9 + 2 \cdot 5\]
\[ = 46\]
Now putting \[x = 24\] and \[y = 0\] in the equation \[Z = 4x + 2y\]
\[Z = 4 \cdot 24 + 2 \cdot 0\]
\[ = 96\]
Now putting \[x = 11.5\] and \[y = 5\] in the equation \[Z = 4x + 2y\]
\[Z = 4 \cdot 11.5 + 2 \cdot 5\]
\[ = 46\]
Thus, the maximum value of 96.
Hence option B is the correct option.

Note: Many students often do a common mistake to calculate the maximum value of \[Z\]. They put intersection point in the equation \[Z = 4x + 2y\] to calculate the maxima value. This is incorrect way. We have to put all corner points in \[Z = 4x + 2y\] to find maximum value of \[Z\].