Ncert Solutions Maths class 12th

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

The maximum value of Z is 32 at (4, 4).

Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits.

Let the factory manufacture x screws of type A and y screws of type B on each day. Therefore,

x ≥ 0 and y ≥ 0

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

The profit on a package of screws A is Rs 7 and on the package of screws B is Rs 10. Therefore, the constraints are

$\begin{array}{l}4x + 6y \le 240\hfill \end{array}$

$\begin{array}{l}6x + 3y \le 240\hfill \end{array}$

Total profit, $Z=7x +10y$

The mathematical formulation of the given problem is

Maximize $Z=7x +10y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}4x + 6y \le 240 ....\left(2\right)\hfill \end{array}$

$\begin{array}{l}6x + 3y \le 240 .....\left(3\right)\hfill \end{array}$

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

The feasible region determined by the system of constraints is

The corner points are A (40, 0), B (30, 20), and C (0, 40).

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

The maximum value of Z is 410 at (30, 20).

Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of ? 410.

Let the manufacturer produce x packages of nuts and y packages of bolts. Therefore,

x ≥ 0 and y ≥ 0

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

The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7. Therefore, the constraints are

$\begin{array}{l}x + 3y \le 12 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + y \le 12 \dots (3\hfill \end{array}$

Total profit, $Z=17.5x +7y$

The mathematical formulation of the given problem is

Maximise $Z=17.5x +7y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}x + 3y \le 12 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + y \le 12 \dots \left(3\right)\hfill \end{array}$

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

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

The corner points are A (4, 0), B (3, 3), and C (0, 4).

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

The maximum value of Z is ? 73.50 at (3, 3).

Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of ? 73.50.

(i) Let the number of rackets and the number of bats to be made be x and y respectively.

The machine time is not available for more than 42 hours.

$\therefore 1.5x+3y\le 42....\left(1\right)$

The craftsman's time is not available for more than 24 hours.

$\therefore 3x+y\le 24 ......\left(2\right)$

The factory is to work at full capacity. Therefore,

$\begin{array}{l}1.5x + 3y = 42\hfill \end{array}$

$\begin{array}{l}3x + y = 24\hfill \end{array}$

On solving these equations, we obtain

x = 4 and y = 12

Thus, 4 rackets and 12 bats must be made.

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

$\begin{array}{l}\therefore 1.5x + 3y \le 42\hfill \end{array}$

$\begin{array}{l}3x + y \le 24\hfill \end{array}$

$\begin{array}{l}x, y \ge 0\hfill \end{array}$

The profit on a racket is Rs 20 and on a bat is Rs 10.

$\therefore Z=20x+10y$

The mathematical formulation of the given problem is

Maximize $Z=20x+10y\dots \left(1\right)$