Ncert Solutions Maths class 12th

Let the mixture contain x kg of food X and y kg of food Y, respectively.

The mathematical formulation of the given problem can be written as given below:

$Minimisez=16x+20y\dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y\ge 10\dots \dots \dots \dots .\left(ii\right)$

$x+y\ge 6\dots \dots \dots \dots \left(iii\right)$

$3x+y\ge 8\dots \dots \dots \dots .\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots \left(v\right)$

The feasible region determined by the system of constraints is given below:

A (10, 0), B (2, 4), C (1, 5) and D (0, 8) are the corner points of the feasible region.

The values of z at these corner points are given below:

Since the feasible region is unbounded, 112 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $16x+20y<112or4x+5y<28$ , and check whether the resulting half-plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with $4x+5y<28$ .

Hence, the minimum value of z is 112 at (2, 4).

Therefore, the mixture should contain 2 kg of food X and 4 kg of food Y.

The minimum cost of the mixture is ? 112.

Let the farmer mix x bags of brand P and y bags of brand Q, respectively

The given information can be compiled in a table as given below:

The given problem can be formulated as given below:

$\begin{array}{}\end{array}Minimisez=250x+200y\dots \dots \dots \dots \left(i\right)$

$3x+1.5y\ge 18\dots \dots \dots \dots ..\left(ii\right)$

$2.5x+11.25y\ge 45\dots \dots \dots ..\left(iii\right)$

$2x+3y\ge 24\dots \dots \dots \dots ..\left(iv\right)$

$x,y\ge 0\dots \dots \dots .\left(v\right)$

The feasible region determined by the system of constraints is given below:

A (18, 0), B (9, 2), C (3, 6) and D (0, 12) are the corner points of the feasible region.

The values of z at these corner points are given below:

Here, the feasible region is unbounded; hence, 1950 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $250x+200y<1950or5x+4y<39$ , and check whether the resulting half-plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with $5x+4y<39$ .

Hence, at point (3, 6), the minimum value of z is 1950.

Therefore, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to ?. 1950.

Let the farmer buy x kg of fertilizer F₁ and y kg of fertilizer F₂. Therefore,

x ≥ 0 and y ≥ 0

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

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.

$\therefore 10\mathrm{\%}of x +5\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{x}{10}+\frac{y}{20}\ge 14\\ 2x+y\ge 280\end{array}$

F₁ consists of 6% phosphoric acid and F₂  consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.

$\therefore 6\mathrm{\%}of x +10\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{6x}{100}+\frac{10y}{100}\ge 14\\ 3x+56y\ge 700\end{array}$

Total cost of fertilizers, $Z=6x +5y$

The mathematical formulation of the given problem is

Minimize $Z=6x +5y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x + y \ge 280 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + 5y \ge 700 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

It can be seen that the feasible region is unbounded.

The corner points are  $A\left(\frac{700}{3},0\right),B\left(100,80\right),C\left(0,280\right)$ .

The values of Z at these points are as follows.

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, $6x +5y <1000$ , and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with

$6x +5y <1000$

Therefore, 100 kg of fertiliser F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is ? 1000.

Let the diet contain x units of food F₁ and y units of food F₂. Therefore,

x ≥ 0 and y ≥ 0

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

The cost of food F₁ is Rs 4 per unit and of Food F₂  is ? 6 per unit. Therefore, the constraints are

$\begin{array}{}\end{array}3x +6y \ge 80$

$4x +3y \ge 100$

$x, y \ge 0$

$Totalcostofthediet,Z=4x +6y$

The mathematical formulation of the given problem is

Minimise $Z=4x +6y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}3x + 6y \ge 80 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}4x + 3y \ge 100 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

The feasible region determined by the constraints is as follows.

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are  $A\left(\frac{8}{3},0\right),B\left(2,\frac{1}{2}\right),C\left(0,\frac{11}{2}\right)$ .

The corner points are $A\left(\frac{80}{3},0\right),B\left(24,\frac{4}{3}\right),C\left(0,\frac{100}{3}\right)$ .

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

As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, $4x +6y <104or2x +3y <52$ , and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with $2x +3y <52$

Therefore, the minimum cost of the mixture will be ? 104.

Let the merchant stock x desktop models and y portable models. Therefore,

x ≥ 0 and y ≥ 0

The cost of a desktop model is Rs 25000 and of a portable model is Rs 4000. However, the merchant can invest a maximum of Rs 70 lakhs.

$\begin{array}{l}\therefore 25000x + 40000y \le 7000000\hfill \end{array}$

$\begin{array}{l}5x +8y \le 1400\hfill \end{array}$

The monthly demand of computers will not exceed 250 units.

$\therefore x+y\le 250$

The profit on a desktop model is Rs 4500 and the profit on a portable model is Rs 5000.

Total profit, $Z=4500x +5000y$

Thus, the mathematical formulation of the given problem is

Maximum $Z=4500x+5000y .....\left(1\right)$

subject to the constraints,

$\begin{array}{l}5x +8y \le 1400 ....\left(2\right)\hfill \end{array}$

$\begin{array}{l}\therefore x + y \le 250 .....\left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 01400 ......\left(4\right)\hfill \end{array}$

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

The corner points are A (250, 0), B (200, 50), and C (0, 175).

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

The maximum value of Z is 1150000 at (200, 50).

Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of ? 1150000.

Let the company manufacture x souvenirs of type A and y souvenirs of type B. Therefore,

x ≥ 0 and y ≥ 0

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

The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6. Therefore, the constraints are

$\begin{array}{l} 5x + 8y \le 200\hfill \end{array}$

$\begin{array}{l}10x + 8y \le 240 i.e.,5x + 4y \le 120\hfill \end{array}$

$\begin{array}{l}Total profit, Z = 5x + 6y\hfill \end{array}$

The mathematical formulation of the given problem is

Maximize $Z=5x +6y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l} 5x + 8y \le 200 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}5x + 4y \le 120 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

The corner points are A (24, 0), B (8, 20), and C (0, 25).

The values of Z at these corner points are as follows

The maximum value of Z is 200 at (8, 20).

Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of ? 160.

Let the cottage industry manufacture x pedestal lamps and y wooden shades. Therefore,

x ≥ 0 and y ≥ 0

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

The profit on a lamp is Rs 5 and on the shades is Rs 3. Therefore, the constraints are

$\begin{array}{l}2x + y \le 12\hfill \end{array}$

$\begin{array}{l}3x + 2y \le 20\hfill \end{array}$

$\begin{array}{l}Total profit, Z = 5x + 3y\hfill \end{array}$

The mathematical formulation of the given problem is

Maximize $Z=5x +3y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x + y \le 12 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + 2y \le 20 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

The corner points are A (6, 0), B (4, 4), and C (0, 10).

The values of Z at these corner points are as follows