Home/Class 12/Maths/

Question and Answer

Show that the minimum of Z occurs at more than two points.
Minimise and maximise \(Z = x + 2y \)subject to \(x + 2 y \geq 100,2 x - y \leq 0,2 x + y \leq 200\) \(x , y \geq 0\)
loading
settings
Speed
00:00
03:22
fullscreen
Show that the minimum of Z occurs at more than two points.
Minimise and maximise \(Z = x + 2y \)subject to \(x + 2 y \geq 100,2 x - y \leq 0,2 x + y \leq 200\) \(x , y \geq 0\)

Answer

see the analysis below
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play

Solution

We have linear constraints as
\(x + 2 y \geq 100\)
\(2 x - y \leq 0\)
\(2x+y\leq 200\)
\(x \geq 0 , y \geq 0\)
and objective function is min or max \((Z) = x + 2y\)
Now, reducing the above inequations into equations and finding their point of intersections i.e.,
\(x + 2y = 100\) ...(i)
\(2x - y = 0\) ...(ii)
\(2x + y = 200\) ...(iii)
\(x = 0, y = 0 \)..(iv)
Solution for Show that the minimum of Z occurs at more than two points.Minimise and maximise Z = x + 2y subject to x + 2 y geq 100,2 x - y leq 0,2 x + y leq 200 x , y geq 0
Now for feasible region, using origin testing method for each constraint
For \(x + 2 y \geq 100\), let \(x = 0, y = 0\)
\(\Rightarrow 0 \geq 100\) i.e., false \(\Rightarrow\) the shaded region will be away from the origin
For \(2 x + y \leq 200\), let \(x = 0, y = 0\)\(\Rightarrow 0 \leq 200\) i.e., true \(\Rightarrow\) The shaded region will be toward the origin.
For \(2 x- y \leq 0\), let \(x = 0, y = 0\)\(\Rightarrow 0 \leq 0\) i.e., true \(\Rightarrow\) The shaded region will be toward the origin.
Also, non-negative restrictions \(x \geq 0 , y \geq 0\) indicates that the feasible region will be exist in first quadrant.
Now, corner points are \(A(0, 50), B(20, 40), C(50, 100)\ and\ D(0, 200)\)
Solution for Show that the minimum of Z occurs at more than two points.Minimise and maximise Z = x + 2y subject to x + 2 y geq 100,2 x - y leq 0,2 x + y leq 200 x , y geq 0
For optimal solution substituting the value of all comer points in \(Z=x+2y\)
Solution for Show that the minimum of Z occurs at more than two points.Minimise and maximise Z = x + 2y subject to x + 2 y geq 100,2 x - y leq 0,2 x + y leq 200 x , y geq 0
Hence, minimum \((Z) = 100\) at all points on the line segment joining the points \(A(0, 50) \)and \(B(20, 40)\) : Maximum \((Z) = 400 \)at \((0, 200)\)
To Keep Reading This Solution, Download the APP
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
To Keep Reading This Solution, Download the APP
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
Correct41
Incorrect0
Watch More Related Solutions
Show that the minimum of Z occurs at more than two points.
Maximize Z = x + y subject to \(x - y \leq - 1, - x + y \leq 0,x,y \geq 0\).
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food
Vitamin A
Vitamin B
Vitamin C
X
1
2
3
Y
2
2
1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?
  • Least cost of the mixture is Rs 122 (2 kg of Food X and 4 kg of food Y).
  • Least cost of the mixture is Rs 132 (3 kg of Food X and 4 kg of food Y).
  • Least cost of the mixture is Rs 112 (2 kg of Food X and 4 kg of food Y).
  • Least cost of the mixture is Rs 142 (2 kg of Food X and 5 kg of food Y).
Solve the following linear programming problem graphically.
Minimize Z = 200 x + 500 y
Subject to constraints
\(x + 2 y \geq 10\)
\(3 x + 4 y \leq 24\)
and \(x \geq 0 , y \geq 0\)
Solve the Linear Programming Problem graphically:
Maximize Z = 3x + 4y subject to the constraints:\(x + y \le4,x \geq 0,y \geq 0\)
A fruit grower can use two types of fertilizer in his garden, brand P and Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brad are given in the table. Testes indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
(Kg per bag)
 Brand PBrand Q
Nitrogen33.5
Phosphoric Acid12
Potash31.5
Chlorine1.52
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A, while each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires at least 240 units of calcium, atleast 460 units of iron and atmost 300 units of cholesterol. How many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Solve the following problem graphically minimize or maximize Z = 3x + 9y
Subject to the constraints :
\(x + 3 y \leq 60\)
\(x + y \geq 10\)
\(x \leq y\)
\(x \geq 0 , y \geq 0\)
Solve the Linear Programming Problem graphically:
Minimize Z = 3x + 5y such that \(x + 3y \geq 3,x + y \geq 2,x,y \geq 0\).
Show that the minimum of Z occurs at more than two points.
Maximize Z = -x + 2y subject to the constraints: \(x \geq 3,x + y \geq 5,x + 2y \geq 6,y \geq 0\).
Solve the Linear Programming Problem graphically:
Minimize Z = -3x + 4y subject to \(x + 2y \leqslant 8,3x + 2y \leqslant 12,x \geqslant 0,y \geqslant 0.\)

Load More