a) Solve the following system of equations by Gauss elimination method. x + 2y = 5 and 2x y = 0 . [2] b) Using simplex method to maximize (Z) = 6x 9y subject to the constraints. x + y 20 ; 2x 3y 6 ;…

a) Solve the following system of equations by Gauss elimination method.

Given: $$\{\begin{array}{l}x+2y=5\text{(1)}\\ 2x-y=0\text{(2)}\end{array}$$

Step 1: Write the augmented matrix: $$\left(\begin{array}{cccc}1& 2& |& 5\\ 2& -1& |& 0\end{array}\right)$$

Step 2: Eliminate $x$ from the second row using $R_2\to R_2-2R_1$: $$\left(\begin{array}{cccc}1& 2& |& 5\\ 0& -5& |& -10\end{array}\right)$$

Step 3: Solve for $y$ from the second row: $$-5y=-10⟹y=2$$

Step 4: Substitute $y=2$ into equation (1): $$x+2(2)=5⟹x=1$$

Solution: $$\boxed{x=1,y=2}$$

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b) Using simplex method to maximize $Z=6x-9y$ subject to:

$$\{\begin{array}{l}x+y\le 20\\ 2x-3y\le 6\\ x\ge 0,y\ge 0\end{array}$$

Step 1: Convert inequalities into equalities by introducing slack variables $s_1,s_2$: $$\{\begin{array}{l}x+y+s_1=20\\ 2x-3y+s_2=6\end{array}$$

Step 2: Write the initial simplex tableau: $$\begin{array}{cccccc}& x& y& s_1& s_2& RHS\\ Z& -6& 9& 0& 0& 0\\ s_1& 1& 1& 1& 0& 20\\ s_2& 2& -3& 0& 1& 6\end{array}$$

Step 3: Identify the entering variable (most negative in $Z$-row): $y$ (coefficient = 9).

Step 4: Compute the ratio test for $y$: $$\frac{20}{1}=20,\frac{6}{-3}=-2(\text{ignore negative})$$ Pivot row: $s_1$ (smallest positive ratio).

Step 5: Perform row operations to make the pivot element 1 and eliminate other entries in the column: $$R_1\to \frac{R_1}{1},R_2\to R_2+3R_1,R_3\to R_3+9R_1$$ New tableau: $$\begin{array}{cccccc}& x& y& s_1& s_2& RHS\\ Z& 3& 0& 9& 0& 180\\ y& 1& 1& 1& 0& 20\\ s_2& -1& 0& 3& 1& 66\end{array}$$

Step 6: No negative coefficients in $Z$-row; optimal solution reached. $$x=0,y=20$$

Maximum value of $Z$: $$\boxed{Z=180}$$

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OR

a) A bullet of mass 0.006 kg travelling at $120{\text{ ms}}^{-1}$ penetrates deeply into a fixed target and is brought to rest in 0.01 sec. Find the distance of penetration.

Given:

• Mass ($m$) = 0.006 kg
• Initial velocity ($u$) = 120 m/s
• Final velocity ($v$) = 0 m/s
• Time ($t$) = 0.01 s

Step 1: Use the equation of motion: $$v=u+at$$ $$0=120+a(0.01)⟹a=-12000{\text{ m/s}}^2$$

Step 2: Use the displacement equation: $$s=ut+\frac{1}{2}a{t}^2$$ $$s=(120)(0.01)+\frac{1}{2}(-12000)(0.01{)}^2$$ $$s=1.2-0.6=0.6\text{ m}$$

Distance of penetration: $$\boxed{0.6\text{ m}}$$

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b) A ball is thrown with the velocity of 29.4 m/sec. Find the two directions in which the ball may be thrown so as to give a range of 44.1 m.

Given:

• Initial velocity ($u$) = 29.4 m/s
• Range ($R$) = 44.1 m
• Acceleration due to gravity ($g$) = 9.8 m/s²

Step 1: Use the range formula: $$R=\frac{u^2\sin (2\theta )}{g}$$ $$44.1=\frac{(29.4{)}^2\sin (2\theta )}{9.8}$$ $$\sin (2\theta )=\frac{44.1\times 9.8}{(29.4{)}^2}=0.5$$

Step 2: Solve for $\theta$: $$2\theta ={\sin }^{-1}(0.5)=30^{\circ }\text{or}150^{\circ }$$ $$\theta =15^{\circ }\text{or}75^{\circ }$$

Two directions: $$\boxed{15^{\circ }\text{ and }75^{\circ }}$$