Question

What is the coordinates of the point at which minimum value of \[z = 7x - 8y\] subject to the constraints \[x + y - 20 \le 0\], \[y \ge 5\], \[x \ge 0\], \[y \ge 0\] is attained, is
A. \[\left( {20,0} \right)\]
B. \[\left( {15,5} \right)\]
C. \[\left( {0,5} \right)\]
D. \[\left( {0,20} \right)\]

Answer 1

Hint: Consider each of the given inequation as an equation. Find the feasible region which satisfy all the given constraints. Then find the coordinates of the corner points of the feasible region by solving the equations. After that find the values of \[z\] at each corner point and observe which is the minimum.

Complete step-by-step solution:
The given function is \[z = 7x - 8y - - - - - \left( i \right)\]
The given inequations are \[x + y - 20 \le 0\], \[y \ge 5\], \[x \ge 0\], \[y \ge 0\]
Let us find the feasible region.
Considering each of the inequations as an equation, we get
\[x + y - 20 = 0 - - - - - \left( {ii} \right)\]
\[y = 5 - - - - - \left( {iii} \right)\]
The constraints \[x \ge 0\] and \[y \ge 0\] indicates the feasible region will be in the first quadrant.
The equation \[\left( {ii} \right)\] represents a straight line which meets \[x\]-axis at the point \[\left( {20,0} \right)\] and \[y\]-axis at the point \[\left( {0,20} \right)\] and the equation \[\left( {iii} \right)\] represents a straight line which is parallel to \[x\]-axis and passes through the point \[\left( {0,5} \right)\].
Putting \[x = 0\] and \[y = 0\] in the first inequation, we get \[ - 20 \le 0\], which is true.
So, the solution region of the first inequation is towards the origin.
Putting \[x = 0\] and \[y = 0\] in the second inequation, we get \[0 \ge 5\], which is false.
So, the solution region of the second inequation is opposite to the origin.
Hence, the feasible region is the triangle \[ABC\].
Coordinates of the vertex \[A\] are \[\left( {0,5} \right)\].
Coordinates of the vertex \[B\] are \[\left( {0,20} \right)\]
Coordinates of the vertex \[C\] is to be determined.
The vertex \[C\] is the point of intersection of the lines \[\left( {ii} \right)\] and \[\left( {iii} \right)\].
Now, solve the equations \[\left( {ii} \right)\] and \[\left( {iii} \right)\] to find the coordinates of the vertex \[C\].
Putting \[y = 5\] in equation \[\left( {ii} \right)\], we get
\[x + 5 - 20 = 0\]
\[ \Rightarrow x - 15 = 0\]
\[ \Rightarrow x = 15\]
Solving equations \[\left( {ii} \right)\] and \[\left( {iii} \right)\], we get \[x = 15\] and \[y = 5\]
So, the coordinates of the vertex \[C\] i.e. the coordinates of the point of intersection of the straight lines \[\left( {ii} \right)\] and \[\left( {iii} \right)\] is \[\left( {15,5} \right)\].
Hence the corner points of the triangle \[ABC\] are \[A\left( {0,5} \right),B\left( {0,20} \right),C\left( {15,5} \right)\].
Now, find the value of \[z = 7x - 8y\] at each of the corner points.
At \[A\left( {0,5} \right)\], \[z = 7 \times 0 - 8 \times 5 = 0 - 40 = - 40\]
At \[B\left( {0,20} \right)\], \[z = 7 \times 0 - 8 \times 20 = 0 - 160 = - 160\]
At \[C\left( {15,5} \right)\], \[z = 7 \times 15 - 8 \times 5 = 105 - 40 = 65\]
The value of \[z\] is minimum at the point \[B\left( {0,20} \right)\].
Hence, option D is correct.

Note: Whenever an inequation is given, you need to consider it as an equation to find the position of the straight line and then put \[x = 0\] and \[y = 0\] in the inequation to indicate the solution region. If the inequation is satisfied by \[x = 0\] and \[y = 0\] then the solution region is in the same direction as that of the origin. After getting the feasible region, you need to find all the corner points. Some corner points may lie at the point of intersection of two straight lines.