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

Step 1: Identify the objective function and constraints.

Maximize $z = 5x + 6y$

Subject to:

Step 2: Find the x- and y-intercepts of the boundary lines.

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: $(0, 2)$.

Step 3: Find the intersection point of $2x + 3y = 10$ and $x + 4y = 8$.

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

Step 4: Identify the corner points of the feasible region.

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.

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

Step 5: Evaluate $z = 5x + 6y$ at each corner point.

Step 6: Determine the maximum value.

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

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

$\boxed{z = 25}$