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Refer to question 13 in short answer question. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?
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$$
\text {Graph of LPP obtained is given as }
$$
Solving $x+y=300$ and $3 x+y=600$, we get $x=150, y=150$
Solving $x-y=-100$ and $x+y=300$, we get
$$
x=100, y=200
$$
And the coordinates of corner points are $\mathrm{O}(0,0)$ $\mathrm{A}(200,0) \mathrm{B}(150,150), \mathrm{C}(100,200)$ and $\mathrm{D}(0,100)$
\begin{array}{|l|l|}
\hline Corner points & Corresponding value of \\
& \mathbf{Z}=\mathbf{2 0 0 x}+\mathbf{1 2 0 y} \\
\hline(0,0) & 0 \\
(200,0) & 40000 \\
(150,150) & 150 \times 200+120 \times 150 \\
& =48000 \leftarrow Maximum \\
(100,200) & 100 \times 200+120 \times 200 \\
& =44000 \\
(0,100) & 120 \times 100=12000 \\
\hline
\end{array}
So maximum profit $=₹ 48000$, when $x=150, y=150$.
\text {Graph of LPP obtained is given as }
$$
Solving $x+y=300$ and $3 x+y=600$, we get $x=150, y=150$
Solving $x-y=-100$ and $x+y=300$, we get
$$
x=100, y=200
$$
And the coordinates of corner points are $\mathrm{O}(0,0)$ $\mathrm{A}(200,0) \mathrm{B}(150,150), \mathrm{C}(100,200)$ and $\mathrm{D}(0,100)$
\begin{array}{|l|l|}
\hline Corner points & Corresponding value of \\
& \mathbf{Z}=\mathbf{2 0 0 x}+\mathbf{1 2 0 y} \\
\hline(0,0) & 0 \\
(200,0) & 40000 \\
(150,150) & 150 \times 200+120 \times 150 \\
& =48000 \leftarrow Maximum \\
(100,200) & 100 \times 200+120 \times 200 \\
& =44000 \\
(0,100) & 120 \times 100=12000 \\
\hline
\end{array}
So maximum profit $=₹ 48000$, when $x=150, y=150$.
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