Class 12th

Let there be x cakes of first kind and y cakes of second kind. Therefore,

x ≥ 0 and y ≥ 0

The given information can be complied in a table as follows.

$\begin{array}{l}\therefore 200x+100y\le 5000\\ \Rightarrow 2x+y\le 50\\ 25x+50y\le 1000\\ \Rightarrow x+2y\le 40\end{array}$

Total numbers of cakes, Z, that can be made are, $Z= x + y$

The mathematical formulation of the given problem is

Maximize $Z= x + y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x+y\le 50.......(2)\\ x+2y\le 40.......(3)\\ x,y\ge 0..............(4)\end{array}$

The feasible region determined by the system of constraints is as follows

The corner points are A (25, 0), B (20, 10), O (0, 0), and C (0, 20).

The values of Z at these corner points are as follows.

Thus, the maximum numbers of cakes that can be made are 30 (20 of one kind and 10 of the other kind).

Maximise $z=x+y$ , subject to $x-y\le -1,-x+y\le 0,x,y\ge 0$

The corresponding equation of the given inequalities are

$x-y=1$ $\frac{x}{-1}+\frac{y}{1}=1$

$-x+y=0$ $\Rightarrow$ $x=y$

$x,y=0$ $x,y=0$

The graph of the given inequalities is shown.

There is no common point in the two shaded region. Thus, there is no feasible region.

$\therefore$ Z has no maximum value.

Maximise $z=-x+2y$ , subject to the constraints

$x\ge 3,x+y\ge 5,x+2y\ge 6,y\ge 0.$

The corresponding equation of the given inequalities are

$x=3$ $x=3$

$x+y=5$ $\Rightarrow \frac{x}{5}+\frac{y}{5}=1$

$x+2y=6$ $\frac{x}{6}+\frac{y}{3}=1$

$y=0$ $y=0$

The graph of the given inequalities is shown

The feasible region unbounded.

The values of Z at corner points A (6,0), B (4,1), and C (3,2) are as follows

As the feasible region is unbounded z=1 may or may not be the maximum values.

So, we plot a graph of $-x+2y>1$

The resulting region has points in common with the feasible region.

Therefore z=1 is not the maximum value. Z has no maximum value.

Minimize and Maximise $z=x+2y$

Subject to $x+2y\ge 100,2x-y\le 0,2x+y\le 200,x,y\ge 0$

The corresponding equation of the given inequalities are

$x+2y=100$ $\Rightarrow \frac{x}{100}+\frac{y}{50}=1$

$2x-y=0$ $2x=y$

$2x+y=200$ $\frac{x}{100}+\frac{y}{200}=1$

$x,y\ge 0$ $x,y=0$

The graph of the inequalities is shown below.

The shaded bounded region ABCD is the feasible region with the corner points.

A (0,50), B, (20,40), C (50,100), D (0,200)

The values of Z at these corner points are

The maximum value of Z is 400 at D (0,200) and the minimum value of Z is 100 at all the points on the line segment joining the points A (0,50) and B (20,40).

Minimize and Maximise $z=5x+10y$

Subject to $x+2y\le 120,x+y\ge 60,x-2y\ge 0,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=120\\ x+y=60\\ x-2y=0\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{120}+\frac{y}{60}=120$

$\begin{array}{l}\frac{x}{60}+\frac{y}{60}=1\\ x=2y\\ x,y=0\end{array}$

The graph of the inequalities in shown.

The shaded founded region ABCD is the feasible region with the corner points $A\left(60,0\right),B\left(120,0\right),C\left(60,30\right)\&D\left(40,20\right)$

The value of Z a these corner points are

The minimum value of Z is 300 at (60,0) and the maximum value of Z is 600 at all the points on the line segment joining B (120,0) and C (60,30).

Minimize $z=x+2y$

Subject to $2x+y\ge 3,x+2y\ge 6,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}2x+y=3\\ x+2y=6\\ x,y\ge 0\end{array}$

$\Rightarrow \frac{x}{\frac{3}{2}}+\frac{y}{3}=1$

$\frac{x}{6}+\frac{y}{3}=1$

$x,y\ge 0$

The feasible region is unbounded the corner point are A (6,0), B (0,3)

The value of Z at these corner points are follows.

Since, the feasible region is unbounded, a graph of $x+2y<6$ is drawn.

Also since there is no point common in feasible region and region $x+2y<6$ .

$z=6$ is maximum on all points joining line (0,3), (6,0)

i.e,  $z=6$ will be minimum on $x+2y=6$ 

Maximum $z=3x+2y$

Subject to $x+2y\le 10,3x+y=\le 15,x,y\ge 0$

The corresponding equation of the given inequalities are :

$\begin{array}{l}x+2y=10\\ 3x+y=15\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{10}+\frac{y}{5}=1$

$\begin{array}{l}\frac{x}{5}+\frac{y}{5}=1\\ x,y=0\end{array}$

The graph of the given inequalities

The shaded bounded region OABC in the feasible region with the corner points

$O(0,0),A(5,0),B(4,3),C(0,5)$

The value of Z at these points are given below.

Therefore, the maximum value of Z is 18 at point (4,3).

Minimize $z=3x+5y$

Such that $x+3y\ge 3,x+y\ge 2,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+3y=3\\ x+y=2\\ x,y=0\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{3}+\frac{y}{1}=1\\ \frac{x}{2}+\frac{y}{2}=1\\ x,y=0\end{array}$

The graph of the given inequalities is

The feasible region is unbounded. The corner points are $A\left(3,0\right),B\left(\frac{3}{2},\frac{1}{2}\right)\&C\left(0,2\right)$

The values of Z at these corner points as follows.

As the feasible region is unbounded, 7 may or may not be minimum value of Z.

We draw the graph of inequality $3x+5y<7$ .

The feasible region has no common point with $3x+5y<7$ .

Therefore minimum value of Z in 7 at $B\left(\frac{3}{2},\frac{1}{2}\right)$

Maximise $z=5x+3y$

Subject to $2x+5y\le 15,5x+2y\le 10,x\ge 0,y\ge 0$

The corresponding equation of the above linear inequalities are

$\begin{array}{l}3x+5y=15\\ 5x+2y=10\&x=0,y=0\end{array}$

$\Rightarrow \frac{x}{5}+\frac{y}{3}=1$

$\begin{array}{l}\frac{x}{2}+\frac{y}{5}=1\\ x=0,y=0\end{array}$

The graph of its given inequalities.

The shaded region OABC is the feasible region which is bounded with the corner points

$O\left(0,0\right),A\left(2,0\right),B\left(\frac{20}{19},\frac{45}{19}\right)\&C\left(0,3\right)$

The values of Z at these points are

Therefore the maximum value of Z is $\frac{235}{19}at,B\left(\frac{20}{19},\frac{45}{19}\right)$