Question Video: Using Linear Optimization to Maximize Profit | Nagwa Question Video: Using Linear Optimization to Maximize Profit | Nagwa

# Question Video: Using Linear Optimization to Maximize Profit Mathematics • First Year of Secondary School

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A factory produces chairs and tables and is trying to decide how many of each it needs to produce to maximize profit. They have determined the constraints and drawn the feasible region as shown, where 𝑥 represents the number of chairs and 𝑦 represents the number of tables. If they find a buyer who agrees to pay a fee such that they receive 150 profit for each chair and 200 profit for each table, what can they expect their maximum profit to be? If they can only guarantee a profit of 50 per chair and 180 per table, how many of each should they produce to maximize their profit?

05:21

### Video Transcript

A factory produces chairs and tables and is trying to decide how many of each it needs to produce to maximize profit. They have determined the constraints and drawn the feasible region as shown, where 𝑥 represents the number of chairs and 𝑦 represents the number of tables. If they find a buyer who agrees to pay a fee such that they receive 150 profit for each chair and 200 profit for each table, what can they expect their maximum profit to be?

Let’s take a look at our feasible region. It’s the region that’s been shaded grey. We know that both 𝑥 and 𝑦 cannot be negative because we couldn’t produce a negative amount of chairs or tables. So we’re only dealing with the first quadrant. And the factory has bounded the other two constraints by these two functions. These constraints produce four extreme points. These extreme points are important because our maximum profit will be found at one of these four points. The feasible region is a quadrilateral. And its four vertices will be the extreme points. The first one is zero, zero. Another one is 45, zero; zero, 32; and 38, 18.

We can go ahead and eliminate the point zero, zero. While zero, zero is a valid extreme point, since the factory’s goal is to maximize their profit, making zero chairs and zero tables would not make them any profit. To find the maximum profit, we need to consider the other three points. Knowing that the profit will be equal to 150 times 𝑥, where 𝑥 is the number of chairs, plus 200 times 𝑦, where 𝑦 is the number of tables.

Let’s plug in what we know. For our first extreme point, the profit will be equal to 150 times 45 plus 200 times zero. 150 times 45 equals 6750, and 200 times zero equals zero. This means, if we made 45 chairs and zero tables, the profit would be 6750. Next, zero, 32, zero chairs and 32 tables, 150 times zero equals zero plus 200 times 32 equals 6400. If the factory made zero chairs and 32 tables, their profit would be 6400. Our last extreme point will consider 150 times 38 plus 200 times 18. This gives us 9300. Making 38 chairs and 18 tables gives us the maximum profit under these conditions for the profit for a chair and a table. So the maximum profit under these conditions is 9300.

Now we want to consider, if they can only guarantee a profit of 50 per chair and 180 per table, how many of each should they produce to maximize their profit? We’ll follow the same procedure. The only thing it’s going to change is the profit for each item. We now wanna consider the profit for a chair being 50 and the profit for a table being 180. It’s still true that we would not maximize our profit by making nothing. Our first spot will consider 45 chairs and zero tables. 50 times 45 plus zero equals a profit of 2250. Moving on to zero chairs and 32 tables, 50 times zero plus 180 times 32 equals a profit of 5760. And our final extreme point 38, 18 equals 5140 of profit.

This time the maximum profit occurs at the extreme point zero, 32. If they can only guarantee 50 per chair and 180 per table, they will make more money if they make zero chairs and 32 tables. Under these conditions, the 𝑥-value should be zero, which means they need zero chairs. The 𝑦-value is 32, which represents 32 tables.

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