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
298.8k+ views
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.
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.
Recently Updated Pages
If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Find the cubic polynomial whose zeroes are 3 5 and class 11 maths JEE_Main

During the sale colour pencils were being sold in -class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In a class of 60 students 25 students play cricket class 11 maths JEE_Main

A regular polygon has 20 sides How many triangles can class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

What Are Current and Potential Difference in Electricity?

