Solved on Feb 23, 2024

Assemble $x$ chain saws and $y$ wood chippers to maximize profit, given 7 hours for chain saw and 2 hours for chipper, with 42 hours total.

STEP 1

Assumptions
1. A chain saw requires 7 hours of assembly time.
2. A wood chipper requires 2 hours of assembly time.
3. There is a maximum of 42 hours of assembly time available.
4. The profit from a chain saw is $150.<br />5. The profit from a wood chipper is$ 240.
6. We need to determine the number of chain saws and wood chippers to assemble for maximum profit.

STEP 2

Let's define variables for the number of chain saws and wood chippers to be assembled. Let $x$ be the number of chain saws and $y$ be the number of wood chippers.

STEP 3

The total assembly time for chain saws and wood chippers cannot exceed 42 hours. This gives us the inequality:
$7x + 2y \leq 42$

STEP 4

The profit function to maximize is given by the profit from each chain saw and wood chipper. The total profit $P$ is:
$P = 150x + 240y$

STEP 5

We need to find the combination of $x$ and $y$ that maximizes $P$ while satisfying the assembly time constraint.

STEP 6

Since we cannot assemble a negative number of chain saws or wood chippers, we have two more inequalities:
$x \geq 0$ $y \geq 0$

STEP 7

To solve this optimization problem, we can use graphical methods or linear programming techniques. For simplicity, we will use the graphical method.

STEP 8

First, we graph the inequality $7x + 2y \leq 42$ on a coordinate plane, where $x$ is on the horizontal axis and $y$ is on the vertical axis.

STEP 9

The boundary line for the inequality is given by $7x + 2y = 42$ . We find the intercepts by setting $x=0$ and $y=0$ in turn.

STEP 10

When $x=0$ :
$7(0) + 2y = 42$ $y = 21$

STEP 11

When $y=0$ :
$7x + 2(0) = 42$ $x = 6$

STEP 12

Plot the points $(0, 21)$ and $(6, 0)$ on the graph and draw the line through them. This line represents all possible combinations of chain saws and wood chippers that use exactly 42 hours of assembly time.

STEP 13

The region that satisfies the inequality $7x + 2y \leq 42$ is below and including the line we just drew.

STEP 14

Plot the intercepts for the inequalities $x \geq 0$ and $y \geq 0$ . These are simply the axes themselves.

STEP 15

The feasible region, where all three inequalities are satisfied, is the area on the graph that is at or above the horizontal axis, at or to the right of the vertical axis, and at or below the line $7x + 2y = 42$ .

STEP 16

The maximum profit will occur at one of the vertices of the feasible region. These vertices are the intercepts we found and the point where the two lines $7x + 2y = 42$ and either $x = 0$ or $y = 0$ intersect.

STEP 17

We already have two vertices: $(0, 21)$ and $(6, 0)$ . We need to check if there is a point of intersection between $7x + 2y = 42$ and the axes within the feasible region.

STEP 18

Since the axes are part of the boundary of the feasible region, the only possible additional vertex is the intersection of $7x + 2y = 42$ with the axes in the first quadrant.

STEP 19

We find this point by solving the system of equations given by $7x + 2y = 42$ and either $x = 0$ or $y = 0$ . We have already found the intercepts, so we know there is no additional vertex.

STEP 20

Now we calculate the profit at each vertex. For $(0, 21)$ :
$P = 150(0) + 240(21) = 5040$

STEP 21

For $(6, 0)$ :
$P = 150(6) + 240(0) = 900$

STEP 22

Since there are no other vertices in the feasible region, we compare the profits at these points. The maximum profit is $5040 at the point$ (0, 21)$.

STEP 23

Therefore, to attain the maximum profit, no chain saws should be assembled and 21 wood chippers should be assembled.
To attain the maximum profit, assemble $0$ chain saws and $21$ wood chippers.

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Assemble xxx chain saws and yyy wood chippers to maximize profit, given 7 hours for chain saw and 2 hours for chipper, with 42 hours total.

STEP 1

STEP 2

Let's define variables for the number of chain saws and wood chippers to be assembled. Let xxx be the number of chain saws and yyy be the number of wood chippers.

STEP 3

The total assembly time for chain saws and wood chippers cannot exceed 42 hours. This gives us the inequality:7x+2y≤427x + 2y \leq 427x+2y≤42

STEP 4

The profit function to maximize is given by the profit from each chain saw and wood chipper. The total profit PPP is:P=150x+240yP = 150x + 240yP=150x+240y

STEP 5

We need to find the combination of xxx and yyy that maximizes PPP while satisfying the assembly time constraint.

STEP 6

Since we cannot assemble a negative number of chain saws or wood chippers, we have two more inequalities:x≥0x \geq 0x≥0 y≥0y \geq 0y≥0

STEP 7

To solve this optimization problem, we can use graphical methods or linear programming techniques. For simplicity, we will use the graphical method.

STEP 8

First, we graph the inequality 7x+2y≤427x + 2y \leq 427x+2y≤42 on a coordinate plane, where xxx is on the horizontal axis and yyy is on the vertical axis.

STEP 9

The boundary line for the inequality is given by 7x+2y=427x + 2y = 427x+2y=42. We find the intercepts by setting x=0x=0x=0 and y=0y=0y=0 in turn.

STEP 10

When x=0x=0x=0:7(0)+2y=427(0) + 2y = 427(0)+2y=42 y=21y = 21y=21

STEP 11

When y=0y=0y=0:7x+2(0)=427x + 2(0) = 427x+2(0)=42 x=6x = 6x=6

STEP 12

Plot the points (0,21)(0, 21)(0,21) and (6,0)(6, 0)(6,0) on the graph and draw the line through them. This line represents all possible combinations of chain saws and wood chippers that use exactly 42 hours of assembly time.

STEP 13

The region that satisfies the inequality 7x+2y≤427x + 2y \leq 427x+2y≤42 is below and including the line we just drew.

STEP 14

Plot the intercepts for the inequalities x≥0x \geq 0x≥0 and y≥0y \geq 0y≥0. These are simply the axes themselves.

STEP 15

The feasible region, where all three inequalities are satisfied, is the area on the graph that is at or above the horizontal axis, at or to the right of the vertical axis, and at or below the line 7x+2y=427x + 2y = 427x+2y=42.

STEP 16

The maximum profit will occur at one of the vertices of the feasible region. These vertices are the intercepts we found and the point where the two lines 7x+2y=427x + 2y = 427x+2y=42 and either x=0x = 0x=0 or y=0y = 0y=0 intersect.

STEP 17

We already have two vertices: (0,21)(0, 21)(0,21) and (6,0)(6, 0)(6,0). We need to check if there is a point of intersection between 7x+2y=427x + 2y = 427x+2y=42 and the axes within the feasible region.

STEP 18

Since the axes are part of the boundary of the feasible region, the only possible additional vertex is the intersection of 7x+2y=427x + 2y = 427x+2y=42 with the axes in the first quadrant.

STEP 19

We find this point by solving the system of equations given by 7x+2y=427x + 2y = 427x+2y=42 and either x=0x = 0x=0 or y=0y = 0y=0. We have already found the intercepts, so we know there is no additional vertex.

STEP 20

Now we calculate the profit at each vertex. For (0,21)(0, 21)(0,21):P=150(0)+240(21)=5040P = 150(0) + 240(21) = 5040P=150(0)+240(21)=5040

STEP 21

For (6,0)(6, 0)(6,0):P=150(6)+240(0)=900P = 150(6) + 240(0) = 900P=150(6)+240(0)=900

STEP 22

Since there are no other vertices in the feasible region, we compare the profits at these points. The maximum profit is 5040atthepoint5040 at the point 5040atthepoint(0, 21)$.

STEP 23

Therefore, to attain the maximum profit, no chain saws should be assembled and 21 wood chippers should be assembled.To attain the maximum profit, assemble 000 chain saws and 212121 wood chippers.

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Assemble $x$ chain saws and $y$ wood chippers to maximize profit, given 7 hours for chain saw and 2 hours for chipper, with 42 hours total.

Let's define variables for the number of chain saws and wood chippers to be assembled. Let $x$ be the number of chain saws and $y$ be the number of wood chippers.

The total assembly time for chain saws and wood chippers cannot exceed 42 hours. This gives us the inequality:
$7x + 2y \leq 42$

The profit function to maximize is given by the profit from each chain saw and wood chipper. The total profit $P$ is:
$P = 150x + 240y$

We need to find the combination of $x$ and $y$ that maximizes $P$ while satisfying the assembly time constraint.

Since we cannot assemble a negative number of chain saws or wood chippers, we have two more inequalities:
$x \geq 0$ $y \geq 0$

First, we graph the inequality $7x + 2y \leq 42$ on a coordinate plane, where $x$ is on the horizontal axis and $y$ is on the vertical axis.

The boundary line for the inequality is given by $7x + 2y = 42$ . We find the intercepts by setting $x=0$ and $y=0$ in turn.

When $x=0$ :
$7(0) + 2y = 42$ $y = 21$

When $y=0$ :
$7x + 2(0) = 42$ $x = 6$

Plot the points $(0, 21)$ and $(6, 0)$ on the graph and draw the line through them. This line represents all possible combinations of chain saws and wood chippers that use exactly 42 hours of assembly time.

The region that satisfies the inequality $7x + 2y \leq 42$ is below and including the line we just drew.

Plot the intercepts for the inequalities $x \geq 0$ and $y \geq 0$ . These are simply the axes themselves.

The feasible region, where all three inequalities are satisfied, is the area on the graph that is at or above the horizontal axis, at or to the right of the vertical axis, and at or below the line $7x + 2y = 42$ .

The maximum profit will occur at one of the vertices of the feasible region. These vertices are the intercepts we found and the point where the two lines $7x + 2y = 42$ and either $x = 0$ or $y = 0$ intersect.

We already have two vertices: $(0, 21)$ and $(6, 0)$ . We need to check if there is a point of intersection between $7x + 2y = 42$ and the axes within the feasible region.

Since the axes are part of the boundary of the feasible region, the only possible additional vertex is the intersection of $7x + 2y = 42$ with the axes in the first quadrant.

We find this point by solving the system of equations given by $7x + 2y = 42$ and either $x = 0$ or $y = 0$ . We have already found the intercepts, so we know there is no additional vertex.

Now we calculate the profit at each vertex. For $(0, 21)$ :
$P = 150(0) + 240(21) = 5040$

For $(6, 0)$ :
$P = 150(6) + 240(0) = 900$

Since there are no other vertices in the feasible region, we compare the profits at these points. The maximum profit is $5040 at the point$ (0, 21)$.

Therefore, to attain the maximum profit, no chain saws should be assembled and 21 wood chippers should be assembled.
To attain the maximum profit, assemble $0$ chain saws and $21$ wood chippers.