Maths M.C.Q (1 Marks)

$y=6, x+y=3, x=0$ and $y=0 y=6$ is the line passing through $(0,6)$ and parallel to the $X$ axis. The region below the line $y=6$ will satisfy the given inequation.

The line $x+y=3$ meets the coordinate axis at $A(3,0)$ and $B(0,3)$. Join these points to obtain the line $x+y=3$.

Clearly, $(0,0)$ satisfies the inequation $x+y \leq 3$. So, the region in $x y$-plane that contains the origin represents the solution set of the given equation.

Region represented by $x \geq 0$ and $y \geq 0$ :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.

The maximum value of $\mathrm{Z}$ is $72 .$

Let the number of cake $A=x$

The number of cake $B=y$

$200 x+100 y \leq 5000$

$\therefore 2 x+y \leq 50$

$25 x+50 y \leq 1000$

$\therefore \quad x+2 y \leq 40$

$x \geq 0, y \geq 0$

$\mathrm{Z}=x+y$

$\therefore$ Equation of the line $\frac{x}{-\frac{3}{4}}+\frac{y}{\frac{3}{2}}=1$

$\therefore-4 x+2 y=3$

$\therefore 4 x-2 y=-3$

Taking $x=y=0 \Rightarrow 0-0 \leq-3$ which is not true.

$\therefore 4 x-2 y \leq-3$ is a half plane not containing $(0,0).$

$(0,0)$

value of $\mathrm{Z}=3 x+2 y$

$\mathrm{Z}=1$

$(1.5,-1.5)$

$Z=1.5$

$(2,0)$

$Z=6$

$(0,2)$

$\mathrm{Z}=4$

$\therefore \mathrm{Z}=3 x+2 y$ has maximum at $\mathrm{C}(2,0)$

At $\mathrm{B}\left(\frac{20}{3}, \frac{20}{3}\right), \mathrm{Z}=x+3 y=\frac{80}{3}$

At $\mathrm{C}(0,10), \mathrm{Z}=x+3 y=30$

$\therefore$ The maximum value of $Z=x+3 y$ is $30.$

For $(3,2), x+2 y=7 \geq 11$ is false.

For $(3,4), x+2 y=11 \geq 11$ is true.

$3 x+4 y=25 \leq 30$ is true.

$2 x+5 y=26 \leq 36$ is true.

 $\therefore$ The objective function is $z=200 x+125 y$.

$\therefore$ The maximum profit is $10,000.$

When $x$ is greater than zero it mean tangent of that mean the feasible region is on or to the right of the line where equation is zero $( x =0)$ which is the line which is $y$ axis

$y \leq 0$

Where $y$ is greater that it means above that mean feasible region is on or above the line where equation is $y =0$ which is the line which is $x$-axis

Those two mean that the feasible region is in the upper right hand part of the xy co ordinate system.

See image $1$

When $x$ is greater that it mean right of

$y \leq 1-x$

That mean feasible region os on or above the line where equation is $x+y=1$

See image $2$

$3 x +2 y \leq 6$

If we solve

$x \leq \frac{6-2 y}{3}$

$y \leq \frac{6-3 x}{2}$

$x$ intercept $(2,0) \quad y \quad$ intercept $(0,3)$

Corner point value of $2=2 x +4 y$

$(1,0)$

$2=2(1)+4(10)=2+0=2$

$(2,0)$

$2=2(2)+4(0)=4+0=4$

$(0,3)$

$2=2(0)+4(3)=0+12=12$

$(0,1)$

$2=2(0)+4(1)=0+4=4$

Maximum value when $x=0 \quad y=3$

$\therefore$ Maximum value of the objective function $300\,<\,325$

$\therefore$ Value of column $(B)$ is maximum.

Maximum value of the objective function $z$ is $12.$

$\therefore$ Maximum value of $z$ of $(3,0)=$ Maximum value of $z=(1,1)$

$\therefore p(3)+q(0)=p(1)+q(1)$

$\therefore p(3)+0=p+q$

$\therefore 3 p=p+q$

$\therefore 2 p=q$

$\therefore p=\frac{q}{2}$

It is given that the maximum value of  $Z$ occurs at two points, $(3,4)$ and $(0,5).$

$\therefore$ Value of $Z$ at $(3,4)=$ Value of $Z$ at $(0,5)$

$\Rightarrow p (3)+ q (4)= p (0)+ q (5)$

$\Rightarrow 3 p+4 q=5 q$

$\Rightarrow q=3 p$

Hence, the correct answer is $B$.

$3 x+4 y \leq 100$  $....(i)$

$4 x+3 y \leq 75$ $......(ii)$

$x \geq 0$

$y \geq 0$

$z \leq \frac{75-3 y}{4}$

$Z=y(6 x+y)$

$Z \leq y\left(6 \cdot\left(\frac{75-3 y}{4}\right)+y\right)$

$z \leq \frac{1}{2}\left(225 y-7 y^{2}\right) \leq \frac{(225)^{2}}{2 \times 4 \times 7}$

$=\frac{50625}{56}$

$\approx 904.0178$

$\approx 904.02$

It will be attained at $y =\frac{225}{14}$

If $x, y$ are optind solutions

So a multiple of  $z=\lambda x+(1-\lambda) y, \lambda \in R$ is abo an optimal sivhin

Since maximum occurs at both $(15,15)$ and $(0,20)$ s, the value $z_0$ is attained at both $(15,15)$ and $(0,20)$

$\Rightarrow z _0= p (15)+ q (15) \text { and } z _0= p (0)+ q (20)$

$\Rightarrow p (15)+ q (15)= p (0)+ q (20)$

$\Rightarrow 15 p =5 q$

$\Rightarrow 3 p = q$

No points on the line $2 x+y=5$ are included.

For $\mathrm{O}\,(0,0), 0+0\,>\,5 \mathrm{which}$ is not true.

$\therefore$ The open half plane not containing origin in the solution set of $2 x+y\,>\,5$

The inequalities $x-y \leq-1$ and $x-y \geq 0$, do not hold together.

$\therefore$ No region exist for the given condition.

Here $x \geq 6 \Rightarrow x \geq 0$

$\therefore x \geq 0$ is not necessary.

$y \geq 2 \Rightarrow y \geq 0 \Rightarrow y \geq 0$ is not necessary.

$2 x+y=2(6)+2=14 \geq 10$

$\therefore 2 x+y \geq 10$ is not necessary.

From the given inequalities, Redundant constraint is $-x+y \geq-10$.

$\therefore$ There are $5$ points.

There are $5$ required points.

Minimum value of the objective function is $-32$

$\therefore$ Minimum value exists at point $(0,8).$

maximum value of objective function $z=3 x-4 y$ is $15$ which exists at point $(5,0).$

we have (Maximum value of $z$ ) $+$ (Minimum value of $z$ ) $=15-32=-17$.

If we take the maximum value of $\mathrm{Z}$ as $42, x_{1}=4, x_{2}=6$

$\therefore 3 x_{1}+2 x_{2}=24 \leq 18$ which is not true.

$x_{1}=2, x_{2}=6,3 x_{1}+2 x_{2}=18 \leq 18$

Also $x_{1} \leq 4, x_{2} \leq 6$

$\therefore$ All conditions are satisfied.

$\therefore$ Maximum value of $\mathrm{Z}$ is $36 .$