Maths

Let godown A supply x and y quintals of grain to shops D and E.

So, $\left(100–x–y\right)$ will be supplied to shop F.

Since x quintals are transported from godown A, the requirement at shop D is 60 quintals. Hence, the remaining (60 – x) quintals will be transported from godown B.

Similarly, (50 – y) quintals and $40–\left(100–x–y\right)=\left(x+y–60\right)$ quintals will be transported from godown B to shop E and F.

The given problem can be represented diagrammatically as given below:

$x\ge 0,y\ge 0,and100–x–y\ge 0$

Then, $x\ge 0,y\ge 0,andx+y\le 100$

$\begin{array}{l}60 – x \ge 0, 50 – y \ge 0, and x + y – 60 \ge 0\hfill \end{array}$

$\begin{array}{l}Then, x \le 60, y \le 50, and x + y \ge 60\hfill \end{array}$

Total transportation cost z is given by,

$\begin{array}{l}z = 6x + 3y + 2.5 \left(100–x–y\right) + 4 \left(60–x\right) + 2 \left(50–y\right) + 3 \left(x+y–60\right)\hfill \end{array}$

$\begin{array}{l}= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180\hfill \end{array}$

$\begin{array}{l}= 2.5x + 1.5y + 410\hfill \end{array}$

The given problem can be formulated as given below:

$Minimisez=2.5x+1.5y+410\dots \dots \dots \dots .\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+y\le 100\dots \dots \dots ..\left(ii\right)$

$x\le 60\dots \dots \dots ..\left(iii\right)$

$y\le 50\dots \dots \dots .\left(iv\right)$

$x+y\ge 60\dots \dots \dots \left(v\right)$

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

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

A (60, 0), B (60, 40), C (50, 50) and D (10, 50) are the corner points.

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

The minimum value of z is 510 at (10, 50).

Hence, the amount of grain transported from A to D, E, and F, is 10 quintals, 50 quintals and 40 quintals, respectively, and from B to D, E and F are 50 quintals, 0 quintals, 0 quintals, respectively.

Thus, the minimum cost is Rs. 510.

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.