QuestionFind the maximum of $p=70x+82y$ subject to: $x \geq 0$ , $y \geq 0$ , $x \leq 10$ , $y \leq 20$ , $x+y \geq 5$ , $x+2y \leq 18$ .

Studdy Solution

STEP 1

Assumptions1. The system of inequalities represents a feasible region in the xy-plane. . The objective function is $p=70x+82y$ .
3. We are looking for the maximum value of the objective function.

STEP 2

The system of inequalities can be represented graphically. Each inequality represents a half-plane. The feasible region is the intersection of these half-planes.

STEP 3

Plot the inequalities on a graph. Start with $x \geq0$ and $y \geq0$ , which represent the first quadrant.

STEP 4

Add the inequality $x \leq10$ , which is a vertical line at $x=10$ . The feasible region is to the left of this line.

STEP 5

Add the inequality $y \leq20$ , which is a horizontal line at $y=20$ . The feasible region is below this line.

STEP 6

Add the inequality $x+y \geq5$ , which is a line passing through the points $(5,0)$ and $(0,5)$ . The feasible region is above this line.

STEP 7

Add the inequality $x+2y \leq18$ , which is a line passing through the points $(9,0)$ and $(0,9)$ . The feasible region is below this line.

STEP 8

The feasible region is the intersection of the regions defined by the inequalities. It is a polygon with vertices at the points of intersection of the lines.

STEP 9

The maximum of the objective function $p=70x+82y$ occurs at one of the vertices of the feasible region. This is because the objective function is linear and the feasible region is a polygon.

STEP 10

Calculate the value of the objective function at each vertex of the feasible region. The vertices are $(0,9)$ , $(5,0)$ , $(10,0)$ , and $(10,20)$ .

STEP 11

Calculate the value of the objective function at $(0,9)$ .
$p =70 \cdot0 +82 \cdot9 =738$

STEP 12

Calculate the value of the objective function at $(5,0)$ .
$p =70 \cdot5 +82 \cdot0 =350$

STEP 13

Calculate the value of the objective function at $(10,0)$ .
$p =70 \cdot10 +82 \cdot0 =700$

STEP 14

Calculate the value of the objective function at $(10,20)$ .
$p =70 \cdot10 +82 \cdot20 =2340$

STEP 15

The maximum value of the objective function is the largest of these values. Therefore, the maximum value of the objective function is $2340$ .
The system has a maximum of $2340$ .

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QuestionFind the maximum of $p=70x+82y$ subject to: $x \geq 0$ , $y \geq 0$ , $x \leq 10$ , $y \leq 20$ , $x+y \geq 5$ , $x+2y \leq 18$ .

Studdy Solution

STEP 1

Assumptions1. The system of inequalities represents a feasible region in the xy-plane. . The objective function is $p=70x+82y$ .
3. We are looking for the maximum value of the objective function.

STEP 2

The system of inequalities can be represented graphically. Each inequality represents a half-plane. The feasible region is the intersection of these half-planes.

STEP 3

Plot the inequalities on a graph. Start with $x \geq0$ and $y \geq0$ , which represent the first quadrant.

STEP 4

Add the inequality $x \leq10$ , which is a vertical line at $x=10$ . The feasible region is to the left of this line.

STEP 5

Add the inequality $y \leq20$ , which is a horizontal line at $y=20$ . The feasible region is below this line.

STEP 6

Add the inequality $x+y \geq5$ , which is a line passing through the points $(5,0)$ and $(0,5)$ . The feasible region is above this line.

STEP 7

Add the inequality $x+2y \leq18$ , which is a line passing through the points $(9,0)$ and $(0,9)$ . The feasible region is below this line.

STEP 8

The feasible region is the intersection of the regions defined by the inequalities. It is a polygon with vertices at the points of intersection of the lines.

STEP 9

The maximum of the objective function $p=70x+82y$ occurs at one of the vertices of the feasible region. This is because the objective function is linear and the feasible region is a polygon.

STEP 10

Calculate the value of the objective function at each vertex of the feasible region. The vertices are $(0,9)$ , $(5,0)$ , $(10,0)$ , and $(10,20)$ .

STEP 11

Calculate the value of the objective function at $(0,9)$ .
$p =70 \cdot0 +82 \cdot9 =738$

STEP 12

Calculate the value of the objective function at $(5,0)$ .
$p =70 \cdot5 +82 \cdot0 =350$

STEP 13

Calculate the value of the objective function at $(10,0)$ .
$p =70 \cdot10 +82 \cdot0 =700$

STEP 14

Calculate the value of the objective function at $(10,20)$ .
$p =70 \cdot10 +82 \cdot20 =2340$

STEP 15

The maximum value of the objective function is the largest of these values. Therefore, the maximum value of the objective function is $2340$ .
The system has a maximum of $2340$ .

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QuestionFind the maximum of p=70x+82yp=70x+82yp=70x+82y subject to: x≥0x \geq 0x≥0, y≥0y \geq 0y≥0, x≤10x \leq 10x≤10, y≤20y \leq 20y≤20, x+y≥5x+y \geq 5x+y≥5, x+2y≤18x+2y \leq 18x+2y≤18.

Studdy Solution

STEP 1

Assumptions1. The system of inequalities represents a feasible region in the xy-plane. . The objective function is p=70x+82yp=70x+82yp=70x+82y.3. We are looking for the maximum value of the objective function.

STEP 2

The system of inequalities can be represented graphically. Each inequality represents a half-plane. The feasible region is the intersection of these half-planes.

STEP 3

Plot the inequalities on a graph. Start with x≥0x \geq0x≥0 and y≥0y \geq0y≥0, which represent the first quadrant.

STEP 4

Add the inequality x≤10x \leq10x≤10, which is a vertical line at x=10x=10x=10. The feasible region is to the left of this line.

STEP 5

Add the inequality y≤20y \leq20y≤20, which is a horizontal line at y=20y=20y=20. The feasible region is below this line.

STEP 6

Add the inequality x+y≥5x+y \geq5x+y≥5, which is a line passing through the points (5,0)(5,0)(5,0) and (0,5)(0,5)(0,5). The feasible region is above this line.

STEP 7

Add the inequality x+2y≤18x+2y \leq18x+2y≤18, which is a line passing through the points (9,0)(9,0)(9,0) and (0,9)(0,9)(0,9). The feasible region is below this line.

STEP 8

The feasible region is the intersection of the regions defined by the inequalities. It is a polygon with vertices at the points of intersection of the lines.

STEP 9

The maximum of the objective function p=70x+82yp=70x+82yp=70x+82y occurs at one of the vertices of the feasible region. This is because the objective function is linear and the feasible region is a polygon.

STEP 10

Calculate the value of the objective function at each vertex of the feasible region. The vertices are (0,9)(0,9)(0,9), (5,0)(5,0)(5,0), (10,0)(10,0)(10,0), and (10,20)(10,20)(10,20).

STEP 11

Calculate the value of the objective function at (0,9)(0,9)(0,9).p=70⋅0+82⋅9=738p =70 \cdot0 +82 \cdot9 =738p=70⋅0+82⋅9=738

STEP 12

Calculate the value of the objective function at (5,0)(5,0)(5,0).p=70⋅5+82⋅0=350p =70 \cdot5 +82 \cdot0 =350p=70⋅5+82⋅0=350

STEP 13

Calculate the value of the objective function at (10,0)(10,0)(10,0).p=70⋅10+82⋅0=700p =70 \cdot10 +82 \cdot0 =700p=70⋅10+82⋅0=700

STEP 14

Calculate the value of the objective function at (10,20)(10,20)(10,20).p=70⋅10+82⋅20=2340p =70 \cdot10 +82 \cdot20 =2340p=70⋅10+82⋅20=2340

STEP 15

The maximum value of the objective function is the largest of these values. Therefore, the maximum value of the objective function is 234023402340.The system has a maximum of 234023402340.

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QuestionFind the maximum of $p=70x+82y$ subject to: $x \geq 0$ , $y \geq 0$ , $x \leq 10$ , $y \leq 20$ , $x+y \geq 5$ , $x+2y \leq 18$ .

Assumptions1. The system of inequalities represents a feasible region in the xy-plane. . The objective function is $p=70x+82y$ .
3. We are looking for the maximum value of the objective function.

Plot the inequalities on a graph. Start with $x \geq0$ and $y \geq0$ , which represent the first quadrant.

Add the inequality $x \leq10$ , which is a vertical line at $x=10$ . The feasible region is to the left of this line.

Add the inequality $y \leq20$ , which is a horizontal line at $y=20$ . The feasible region is below this line.

Add the inequality $x+y \geq5$ , which is a line passing through the points $(5,0)$ and $(0,5)$ . The feasible region is above this line.

Add the inequality $x+2y \leq18$ , which is a line passing through the points $(9,0)$ and $(0,9)$ . The feasible region is below this line.

The maximum of the objective function $p=70x+82y$ occurs at one of the vertices of the feasible region. This is because the objective function is linear and the feasible region is a polygon.

Calculate the value of the objective function at each vertex of the feasible region. The vertices are $(0,9)$ , $(5,0)$ , $(10,0)$ , and $(10,20)$ .

Calculate the value of the objective function at $(0,9)$ .
$p =70 \cdot0 +82 \cdot9 =738$

Calculate the value of the objective function at $(5,0)$ .
$p =70 \cdot5 +82 \cdot0 =350$

Calculate the value of the objective function at $(10,0)$ .
$p =70 \cdot10 +82 \cdot0 =700$

Calculate the value of the objective function at $(10,20)$ .
$p =70 \cdot10 +82 \cdot20 =2340$

The maximum value of the objective function is the largest of these values. Therefore, the maximum value of the objective function is $2340$ .
The system has a maximum of $2340$ .