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The University of Ottawa and McMaster University are some of the best universities in the Canada. Both the universities have their own advantages and disadvantages. According to Shiksha Popularity Rankings, University of Ottawa has an overall ranking of #220 in World. Whereas, the McMaster University has an overall ranking of #152 in World. In comparison of Shiksha Popularity Rankings, McMaster University stands taller as compared to Ottawa University. Check the comparison listed below between University of Ottawa and McMaster University. The below comparison will surely help students in gaining more clarity on these universities respectively:

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.

The University of Ottawa and Queens University are some of the most reputed universities in Canada. Both the universities are among the top choices of international students. The University of Ottawa is located in Ottawa, Ontario, Canada and Queens University is located in Kingston, Ontario, Canada. University of Ottawa acceptance rate stands at 42%, which makes it highly competitive. This figure indicates that out of every 100 students, only 42 students enrolled at the university. Whereas, Queens University acceptance rate is around 64%, which makes it competitive. This means out of 100 students around 64 students are enrolled at the university. Check the comparison listed below between University of Ottawa and Queens University. This will surely help internationals students to gain clarity on both the universities:

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.

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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.