Question

$\underline{\huge\color{orange}\bf QUESTION}$
Minimize z = 3x + 5y subject to the constraints x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0​

Answer 1

$\large\underline{\sf{Solution-}}$

We first draw the lines,

• x + 3y = 3, x + y = 2

Consider,

$\red{\bf :\longmapsto\:x + 3y = 3 - - - (1)}$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + 3y = 3$

$\bf\implies \:y = 1$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x + 3 \times 0 = 3$

$\bf\implies \:x = 3$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 1) & (3 , 0)

➢ See the attachment graph.

Now,

Consider,

$\blue{\bf :\longmapsto\:x + y = 2 - - - (2)}$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + y = 2$

$\bf\implies \:y = 2$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x + 0 = 2$

$\bf\implies \:x = 2$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 2 \\ \\ \sf 2 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 2)& (2 , 0)

➢ See the attachment graph.

Now, shade the region satisfied the given inequalities.

We observe that the feasible region is unbounded and having coordinates,

• B (3, 0)

• D (0, 2)

• E (1.5, 0.5)

Now,

• At the corner points, the value of Z is

$\begin{gathered}\boxed{\begin{array}{c|c} \bf points & \bf value \: of \: Z = 3x + 5y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf D (0, 2) & \sf 10 \\ \\ \sf B (3, 0) & \sf 9\\ \\ \sf E (1.5, 0.5) & \sf 7 \end{array}} \\ \end{gathered}$

Thus,

• Minimum value of Z = 7 at E (1.5, 0.5)

As the feasible region is unbounded, we are not sure whether this is the minimum value or not.

Justification :-

Consider the half plane Z < 7

It means

• 3x + 5y < 7

We now sketch this inequality and found that there is no common point with the feasible region.

Hence,

• Z = 7 is the minimum value at E (1.5, 0.5)

Answer 2

$\boxed {\boxed{ { \green{ \bold{ \underline{Verified \: Answer \: }}}}}}$

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$\bigstar{\underline{\underline{{\sf\ \red{ Solution-}}}}}$

Zmin. = 3x + 5y

x + 3y ≥ 3

x + y ≥ 2

where x ≥ 0, y ≥ 0

Note:- Full answer attached in image

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