Q1. A binary operation * is defined on the set of real numbers R, by P*q = P+2q. Evaluate 5/7 * 4/3. If (x+3)*3 = -5, find the value of x.

Maximize $z = 5x + 6y$

Subject to:

$$\begin{cases} 2x + 3y \leq 10 \\ x + 4y \leq 8 \\ x \geq 0 \\ y \geq 0 \end{cases}$$

For $2x + 3y = 10$:

• x-intercept: $y = 0$, $2x = 10$, $x = 5$. Point: $(5, 0)$.

• y-intercept: $x = 0$, $3y = 10$, $y = \frac{10}{3}$. Point: $\left(0, \frac{10}{3}\right)$.

For $x + 4y = 8$:

• x-intercept: $y = 0$, $x = 8$. Point: $(8, 0)$.

• y-intercept: $x = 0$, $4y = 8$, $y = 2$. Point: $(0, 2)$.

Solve the system:

$$x + 4y = 8 \quad (1)$$ $$2x + 3y = 10 \quad (2)$$

Multiply (1) by 2: $2x + 8y = 16 \quad (3)$

Subtract (2) from (3): $(2x + 8y) - (2x + 3y) = 16 - 10$

$$5y = 6$$ $$y = \frac{6}{5}$$

Substitute into (1): $x + 4\left(\frac{6}{5}\right) = 8$

$$x + \frac{24}{5} = 8$$ $$x = 8 - \frac{24}{5} = \frac{40}{5} - \frac{24}{5} = \frac{16}{5}$$

Intersection point: $\left(\frac{16}{5}, \frac{6}{5}\right)$.

The corner points are: $(0,0)$, $(5,0)$, $\left(\frac{16}{5}, \frac{6}{5}\right)$, $(0,2)$.

Verify feasibility:

• $(0,0)$: $2(0)+3(0)=0 \leq 10$, $0+4(0)=0 \leq 8$.

• $(5,0)$: $2(5)+3(0)=10 \leq 10$, $5+4(0)=5 \leq 8$.

• $\left(\frac{16}{5}, \frac{6}{5}\right)$: On both lines, feasible.

• $(0,2)$: $2(0)+3(2)=6 \leq 10$, $0+4(2)=8 \leq 8$.

The maximum $z = 25$ occurs at $(5, 0)$.

Optimal solution: $z = 25$ at $x=5, y=0$.