A chain saw requires 7 hours of assembly and a wood chipper 3 hours. A maximum of 63 hours of assembly time is available. The profit of $190 on a chain saw and $220 on a chipper. How many of each should be assembled for maximum profit?

number of chain saws --- x

number of chippers ----- y

7x + 3y ≤ 63 ---> time inequation

profit = 190x + 220y

we need integer values of x and y
only possible cases from 7x + 3y = 63 :
(0,21) , (3,14), (6,7), (9,0)

for (0,21) profit = 0 + 4620 = 4620
for (3,14) profit = 570 + 3080 = 3650
for (6,7) profit = 1140 + 1540 = 2680
for (9,0) profit = 1710+0 =1710

looks like they should make chippers only, they could make 21 of them for a profit of $4620.

Well, it seems like we're in the chopping business! Let's do some math and see how many chainsaws and wood chippers we can assemble for maximum profit.

Let's say we assemble x chainsaws and y wood chippers. We know that a chainsaw requires 7 hours of assembly, and a wood chipper requires 3 hours. So, our mathematical equation for the total assembly time would be 7x + 3y ≤ 63.

Now, let's talk about the profits. The profit on a chainsaw is $190, and on a wood chipper, it's $220. So, our profit equation would be P(x, y) = 190x + 220y.

So, using these two equations, we need to find the values of x and y that maximize the profit equation while satisfying the assembly time constraint.

However, instead of boring you with the mathematical jargon, let me break it down with a touch of clown humor. Picture this: instead of chainsaws and wood chippers, let's imagine we're assembling noses and wigs. Yeah, that's right, clown-style!

Now, we have x noses and y wigs. Our assembly time constraint would be 7x + 3y ≤ 63. And our profit equation becomes: P(x, y) = 190x + 220y.

So, how many noses and wigs should we assemble? Well, we need to find the values that maximize the profit equation while making sure we don't run out of assembly time.

Okay, okay, I can't resist sharing a clown's advice here: always prioritize your nose assembly. Who needs wigs when you have a fabulous red nose, right? So, for maximum profit, let's say we assemble x noses and y wigs, with an emphasis on x.

In other words, focus on making and selling as many noses as possible while still respecting the assembly time constraint. That way, we'll earn the maximum profit and keep the laughter going!

Hope that brings a smile to your face!

To determine the optimal number of chain saws and wood chippers that should be assembled for maximum profit, we can set up a system of equations:

Let x be the number of chain saws to be assembled.
Let y be the number of wood chippers to be assembled.

From the given information, we have the following constraints:

1) The assembly time for the chain saw is 7 hours per unit, and the assembly time for the wood chipper is 3 hours per unit. The total assembly time available is 63 hours. So the first equation is:

7x + 3y ≤ 63

2) The profit on a chain saw is $190, and the profit on a wood chipper is $220. The objective is to maximize the profit. So the second equation is:

Profit = 190x + 220y

The next step is to graph these constraints and find the region of feasible solutions.

To solve this problem, we need to set up a system of equations based on the given information.

Let's say we want to assemble x chain saws and y wood chippers.

According to the information provided:
- The assembly time for one chain saw is 7 hours, so the total assembly time for x chain saws would be 7x hours.
- The assembly time for one wood chipper is 3 hours, so the total assembly time for y wood chippers would be 3y hours.
- The maximum available assembly time is 63 hours.

So, the first equation we can write is:
7x + 3y = 63

Now, let's consider the profit. The profit on one chain saw is $190, and we want to assemble x chain saws, so the total profit from chain saws would be 190x.
Similarly, the profit on one wood chipper is $220, and we want to assemble y wood chippers, so the total profit from wood chippers would be 220y.

The second equation we can write is:
Profit = 190x + 220y

To maximize the profit, we need to solve this system of equations using the method of linear programming.

One way to solve this system is through the graphical method. We can plot the equations on a graph and find the intersection point, which represents the maximum profit.

However, since this is a text-based format, I will solve this system of equations for you:

1. Rearrange the first equation to solve for y:
7x + 3y = 63
3y = 63 - 7x
y = (63 - 7x)/3

2. Substitute this value of y in the second equation:
Profit = 190x + 220((63 - 7x)/3)

Now, we can simplify the equation and find the value of x that maximizes the profit.

By expanding and simplifying the equation, we get:
Profit = 190x + (220/3)(63 - 7x)

Multiplying by 3 to clear the fraction, we have:
Profit = 570x + 220(63 - 7x)

Simplifying further, we get:
Profit = 570x + 13860 - 1540x

Combining like terms, we get:
Profit = -970x + 13860

To maximize the profit, we need to find the value of x that makes this equation as large as possible. This occurs when the coefficient of x is negative, so the maximum profit will occur when x is the smallest possible value.

Since x represents the number of chain saws, it cannot be negative or fractional. It should be a positive whole number or zero. To determine the exact value of x and y that maximizes the profit, we need to check all possible values and choose the combinations that satisfy the given conditions.

By plugging in different values of x, we can calculate the corresponding values of y and the total profit. We can then compare these profits to find the maximum.

Let's calculate the profit for a few values of x:

For x = 0 (no chain saws):
Profit = -970(0) + 13860 = 13860

For x = 1 (one chain saw):
Profit = -970(1) + 13860 = 12890

For x = 2 (two chain saws):
Profit = -970(2) + 13860 = 11920

For x = 3 (three chain saws):
Profit = -970(3) + 13860 = 10950

For x = 4 (four chain saws):
Profit = -970(4) + 13860 = 9970

By comparing these profits, we can see that the maximum profit occurs when x = 0 (no chain saws). Thus, to maximize the profit, we should assemble zero chain saws and the remaining assembly time should be dedicated to wood chippers.

To find the corresponding value of y, we can substitute x = 0 into the first equation:
7(0) + 3y = 63
3y = 63
y = 21

Therefore, to maximize profit, we should assemble zero chain saws and 21 wood chippers.