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  • Maximize the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.

Maximize the function Z = 11x + 7y, subject to the constraints: x \leq 3, y \leq 2, x \geq 0, y \geq 0.\\

Answers (1)

It is subject to constraints

x $ \leq $ 3, y $ \leq $ 2, x $ \geq $ 0, y $ \geq $ 0.\\

Now let us convert the given inequalities into equation
We obtain the following equation

\\ x \leq 3 \\ \Rightarrow x=3 \\ y \leq 2 \\ \Rightarrow y=2 \\ y \geq 0 \\ \Rightarrow x=0 \\ y=0

The region represented by x≤3:

The line is parallel to the Y-axis and then meets the X-axis at the point x=3. Then it further gives a clarification that it satisfies the inequation in the problem that isx$ \leq $ 2. The region then represents the origin and the set of the inequation x$ \leq $ 3.\\

The region that is represented by y$ \leq $ 2:\\

The line that is parallel to the x-axis meets the y-axis. The part that contains the region represents the solution set of the other inequation y$ \leq $ 2.\\

Therefore, the region that represents the x$ \geq $ 0 \ and\ y$ \geq $ 0 is first quadrant and satisfies the inequations. After plotting the graph we get 

 

The shaded region OBCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,2), \mathrm{C}(3,2)$ and $\mathrm{D}(3,0)$
Now we will substitute these values in Z at each of these corner points, we get

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline O(0,0) & Z=11(0)+7(0)=0+0=0 \\ B(0,2) & Z=11(0)+7(2)=0+14=14 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \rightarrow \max \\ D(3,0) & Z=11(3)+7(0)=33+0=33 \\ \hline \end{array}

Therefore, the final answer is that the value of Z is 47 at the point of (3,2).

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