maths

Let x be the no. of nuts & y be the bolts.

We have to maximize x & y if the factory worked at capacity of 12hours per day.

Clearly, $x≥y,y≥0$

Since we have only 12 hours of machine, we will use the following constraints:-

$x+3y≤12$

$3x+y≤12$

The profit of nuts is Rs.17.5 and bolts is Rs. 7 .

We need to maximize $17.5x+7y$

We can see that the feasible region is bounded and in the first quadrant.

On solving the equations, $x+3y=12$ and $3x+y=12$, we get $x=3$ and $y=3$.

Therefore, feasible points are $(0,0);(0,4);(3,3);(4,4)$

Values of z are($z=17.5x+7y$)

The maximum profit $=73.5$ which is when $3$ packets each of nuts & bolts are produced.

Answered By

toppr

How satisfied are you with the answer?

This will help us to improve better

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

View Answer

VIEW MORE

ExampleDefinitionsFormulaes