Question

In Andrew’s Furniture Shop, he builds bookshelves and tables. Each type of furniture takes him about the same time to make. He figures he has time to make, at most, 25 pieces of furniture by this Saturday. The materials for each bookshelf cost him $20.00 and the materials for each table cost him $45.00. He has $675.00 to spend on materials. Andrew makes a profit of $60.00 on each bookshelf and a profit of $100.00 for each table. How many of each piece of furniture should Andrew make to maximize profit? There should be 4 inequalites. I am not sure on how to make the optimization or objective equation

Answer 1

Answer:

Andrew should make 7 tables and 18 bookshelves.

Explanation:

Let the number of tables made = x

Let the number of bookshelves made = y

Therefore:

$x\geq 0\\y \geq 0$

Andrew has time to make, at most, 25 pieces of furniture by this Saturday.

$x+y\leq 25$

The materials for each table cost him $45.00.

The materials for each bookshelf cost him $20.00

Total Cost = 45x+20y

Since has $675.00 to spend on materials.

$45x+20y\leq 675$

He makes a profit of $60.00 on each bookshelf and a profit of $100.00 for each table.

Therefore, the profit function, P(x,y)=100x+60y

Therefore, the required problem is:

Objective Function: Max P(x,y)=100x+60y

Subject to the constraints:

$45x+20y\leq 675$

$x+y\leq 25$

$x\geq 0\\y \geq 0$

The graph is plotted and the vertices of the feasible region are:

(0,25), (0,0), (7,18) and (15,0).

At (0,25), P(x,y)=100x+60y=100(0)+60(25)=$1500

At (0,0), P(x,y)=100x+60y=100(0)+60(0)=$0

At (7,18), P(x,y)=100x+60y=100(7)+60(18)=$1780

At (15,0), P(x,y)=100x+60y=100(15)+60(0)=$1500

We see that P(x,y) has a maximum at (7,18).

Therefore, to make maximum profit, Andrew should make 7 tables and 18 bookshelves.