Linear Programming

Maximize $W=5x+2y$ subject to $3x+5y\le15,\ 5x+2y\le10,\ x,y\ge0$.

Add slack variables $s_1,s_2\ge0$:

$3x+5y+s_1=15,\qquad 5x+2y+s_2=10$

Initial table

The largest positive $C_j-Z_j$ is 5, so $x$ enters; the minimum ratio is 2, so $s_2$ leaves. Pivot element = 5.

Iteration 1: $R_2\to\frac15R_2$, then $R_1\to R_1-3R_2$.

All $C_j-Z_j\le0$, so the solution is optimal.

$x=2,\quad y=0,\quad W_{\max}=5(2)+2(0)=10$

The maximum value of $W$ is 10, at $x=2$, $y=0$.

Convert the minimization problem into a maximization problem:

$$\text{Maximize } Z=-P=-3x-4y$$

Subject to

$$2x+y\le 4,\qquad x+2y\le 4,\qquad x,y\ge 0$$

Introduce slack variables $s_1,s_2\ge 0$:

$$2x+y+s_1=4$$ $$x+2y+s_2=4$$

Initial simplex tableau:

For maximization, the current solution is optimal if all entries in the $C_j-Z_j$ row are $\le 0$.

Here,

$$C_j-Z_j=(-3,-4,0,0)$$

All are $\le 0$. So no further iteration is needed.

The optimal solution is

$$x=0,\qquad y=0$$

and

$$Z=0$$

Since $Z=-P$,

$$P=-Z=0$$

Therefore, the minimum value is

$$\boxed{P_{\min}=0}$$

at

$$\boxed{x=0,\ y=0}$$

Maximise and minimise $F=10x+15y$ subject to $x+2y\le20,\quad x+y\le16,\quad x\ge0,\ y\ge0.$

Boundary lines:

• $x+2y=20$ passes through $(20,0)$ and $(0,10)$.
• $x+y=16$ passes through $(16,0)$ and $(0,16)$.

Intersection of the two lines: subtracting $x+y=16$ from $x+2y=20$ gives $y=4$, so $x=12$. The point is $(12,4)$.

Feasible region: the region bounded by the axes and the two lines (on the origin side of both). Its corner points are $O(0,0),\quad A(16,0),\quad B(12,4),\quad C(0,10).$

Value of $F$ at each corner:

The extreme values of a linear function over a bounded feasible region occur at corner points.

$\boxed{F_{\max}=180\text{ at }(12,4),\qquad F_{\min}=0\text{ at }(0,0)}$

(c) 0.5

Hawkin–Simon condition requires all principal minors to be positive.
Here the second diagonal entry is $0.3>0$, and the determinant must satisfy

$$0.3x-(-0.1)(-0.2)=0.3x-0.02>0$$

so $x>0.0667$. Among the options, only $0.5$ satisfies this.

With $m$ equations and $n$ unknowns ($n>m$), a basic solution is found by setting $n-m$ variables to zero and solving for the other $m$. The number of basic solutions is therefore

$\binom{n}{m}=\frac{n!}{m!\,(n-m)!}$

None of the options (a)-(d) equals this, so the question appears to have a missing or mistyped option.