Solved on Dec 06, 2023

Minimize $c=x+2y$ subject to $x+5y \geq 19, 4x+y \geq 19, x \geq 0, y \geq 0$ , where $c = (x,y)=(\sqrt{x})$ .

STEP 1

Assumptions
1. The objective function to minimize is $c = x + 2y$ .
2. The constraints are given by the inequalities: $x + 5y \geq 19$ , $4x + y \geq 19$ , $x \geq 0$ , $y \geq 0$ .
3. The problem is a linear programming problem.

STEP 2

Draw the feasible region defined by the constraints on a coordinate plane.
- The first constraint $x + 5y \geq 19$ represents a line when $x + 5y = 19$ . We can find two points on this line by setting $x = 0$ and $y = 0$ and solving for the other variable. - The second constraint $4x + y \geq 19$ represents another line when $4x + y = 19$ . We can find two points on this line in the same manner.

STEP 3

Find the point on the line $x + 5y = 19$ when $x = 0$ .
$y = \frac{19}{5}$

STEP 4

Find the point on the line $x + 5y = 19$ when $y = 0$ .
$x = 19$

STEP 5

Find the point on the line $4x + y = 19$ when $x = 0$ .
$y = 19$

STEP 6

Find the point on the line $4x + y = 19$ when $y = 0$ .
$x = \frac{19}{4}$

STEP 7

Plot the lines and the feasible region on a coordinate plane. The feasible region is the area that satisfies all inequalities, including $x \geq 0$ and $y \geq 0$ .

STEP 8

Identify the corner points of the feasible region. These are the points where the constraint lines intersect each other and the axes.

STEP 9

Find the intersection point of the lines $x + 5y = 19$ and $4x + y = 19$ by solving the system of equations.

STEP 10

Set up the system of equations:
$\begin{align} x + 5y &= 19 \\ 4x + y &= 19 \end{align}$

STEP 11

Multiply the second equation by 5 to eliminate $y$ :
$\begin{align} x + 5y &= 19 \\ 20x + 5y &= 95 \end{align}$

STEP 12

Subtract the first equation from the second equation to solve for $x$ :
$\begin{align} (20x + 5y) - (x + 5y) &= 95 - 19 \\ 19x &= 76 \\ x &= 4 \end{align}$

STEP 13

Substitute $x = 4$ into the first equation to solve for $y$ :
$\begin{align} 4 + 5y &= 19 \\ 5y &= 15 \\ y &= 3 \end{align}$

STEP 14

The intersection point of the two lines is $(4, 3)$ .

STEP 15

Evaluate the objective function $c = x + 2y$ at each corner point of the feasible region, including $(4, 3)$ , $(19, 0)$ , $(0, \frac{19}{5})$ , and $(0, 19)$ .

STEP 16

Calculate $c$ at $(4, 3)$ :
$c = 4 + 2(3) = 4 + 6 = 10$

STEP 17

Calculate $c$ at $(19, 0)$ :
$c = 19 + 2(0) = 19$

STEP 18

Calculate $c$ at $(0, \frac{19}{5})$ :
$c = 0 + 2\left(\frac{19}{5}\right) = \frac{38}{5} = 7.6$

STEP 19

Calculate $c$ at $(0, 19)$ :
$c = 0 + 2(19) = 38$

STEP 20

The minimum value of $c$ is the smallest value obtained from the corner points.

STEP 21

Compare the values of $c$ to find the minimum:
- At $(4, 3)$ , $c = 10$ - At $(19, 0)$ , $c = 19$ - At $(0, \frac{19}{5})$ , $c = 7.6$ - At $(0, 19)$ , $c = 38$

STEP 22

The minimum value of $c$ is $7.6$ at the point $(0, \frac{19}{5})$ .
The solution to the problem is that the minimum value of the objective function $c$ is $7.6$ , and it occurs when $x = 0$ and $y = \frac{19}{5}$ .

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Minimize c=x+2yc=x+2yc=x+2y subject to x+5y≥19,4x+y≥19,x≥0,y≥0x+5y \geq 19, 4x+y \geq 19, x \geq 0, y \geq 0x+5y≥19,4x+y≥19,x≥0,y≥0, where c=(x,y)=(x)c = (x,y)=(\sqrt{x})c=(x,y)=(x​).

STEP 1

Assumptions1. The objective function to minimize is c=x+2y c = x + 2y c=x+2y.2. The constraints are given by the inequalities: x+5y≥19 x + 5y \geq 19 x+5y≥19, 4x+y≥19 4x + y \geq 19 4x+y≥19, x≥0 x \geq 0 x≥0, y≥0 y \geq 0 y≥0.3. The problem is a linear programming problem.

STEP 2

STEP 3

Find the point on the line x+5y=19 x + 5y = 19 x+5y=19 when x=0 x = 0 x=0.y=195 y = \frac{19}{5} y=519​

STEP 4

Find the point on the line x+5y=19 x + 5y = 19 x+5y=19 when y=0 y = 0 y=0.x=19 x = 19 x=19

STEP 5

Find the point on the line 4x+y=19 4x + y = 19 4x+y=19 when x=0 x = 0 x=0.y=19 y = 19 y=19

STEP 6

Find the point on the line 4x+y=19 4x + y = 19 4x+y=19 when y=0 y = 0 y=0.x=194 x = \frac{19}{4} x=419​

STEP 7

Plot the lines and the feasible region on a coordinate plane. The feasible region is the area that satisfies all inequalities, including x≥0 x \geq 0 x≥0 and y≥0 y \geq 0 y≥0.

STEP 8

Identify the corner points of the feasible region. These are the points where the constraint lines intersect each other and the axes.

STEP 9

Find the intersection point of the lines x+5y=19 x + 5y = 19 x+5y=19 and 4x+y=19 4x + y = 19 4x+y=19 by solving the system of equations.

STEP 10

Set up the system of equations:x+5y=194x+y=19 \begin{align*} x + 5y &= 19 \\ 4x + y &= 19 \end{align*} x+5y4x+y​=19=19​

STEP 11

Multiply the second equation by 5 to eliminate y y y:x+5y=1920x+5y=95 \begin{align*} x + 5y &= 19 \\ 20x + 5y &= 95 \end{align*} x+5y20x+5y​=19=95​

STEP 12

Subtract the first equation from the second equation to solve for x x x:(20x+5y)−(x+5y)=95−1919x=76x=4 \begin{align*} (20x + 5y) - (x + 5y) &= 95 - 19 \\ 19x &= 76 \\ x &= 4 \end{align*} (20x+5y)−(x+5y)19xx​=95−19=76=4​

STEP 13

Substitute x=4 x = 4 x=4 into the first equation to solve for y y y:4+5y=195y=15y=3 \begin{align*} 4 + 5y &= 19 \\ 5y &= 15 \\ y &= 3 \end{align*} 4+5y5yy​=19=15=3​

STEP 14

The intersection point of the two lines is (4,3) (4, 3) (4,3).

STEP 15

Evaluate the objective function c=x+2y c = x + 2y c=x+2y at each corner point of the feasible region, including (4,3) (4, 3) (4,3), (19,0) (19, 0) (19,0), (0,195) (0, \frac{19}{5}) (0,519​), and (0,19) (0, 19) (0,19).

STEP 16

Calculate c c c at (4,3) (4, 3) (4,3):c=4+2(3)=4+6=10 c = 4 + 2(3) = 4 + 6 = 10 c=4+2(3)=4+6=10

STEP 17

Calculate c c c at (19,0) (19, 0) (19,0):c=19+2(0)=19 c = 19 + 2(0) = 19 c=19+2(0)=19

STEP 18

Calculate c c c at (0,195) (0, \frac{19}{5}) (0,519​):c=0+2(195)=385=7.6 c = 0 + 2\left(\frac{19}{5}\right) = \frac{38}{5} = 7.6 c=0+2(519​)=538​=7.6

STEP 19

Calculate c c c at (0,19) (0, 19) (0,19):c=0+2(19)=38 c = 0 + 2(19) = 38 c=0+2(19)=38

STEP 20

The minimum value of c c c is the smallest value obtained from the corner points.

STEP 21

Compare the values of c c c to find the minimum:- At (4,3) (4, 3) (4,3), c=10 c = 10 c=10 - At (19,0) (19, 0) (19,0), c=19 c = 19 c=19 - At (0,195) (0, \frac{19}{5}) (0,519​), c=7.6 c = 7.6 c=7.6 - At (0,19) (0, 19) (0,19), c=38 c = 38 c=38

STEP 22

The minimum value of c c c is 7.6 7.6 7.6 at the point (0,195) (0, \frac{19}{5}) (0,519​).The solution to the problem is that the minimum value of the objective function c c c is 7.6 7.6 7.6, and it occurs when x=0 x = 0 x=0 and y=195 y = \frac{19}{5} y=519​.

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Minimize $c=x+2y$ subject to $x+5y \geq 19, 4x+y \geq 19, x \geq 0, y \geq 0$ , where $c = (x,y)=(\sqrt{x})$ .

Assumptions
1. The objective function to minimize is $c = x + 2y$ .
2. The constraints are given by the inequalities: $x + 5y \geq 19$ , $4x + y \geq 19$ , $x \geq 0$ , $y \geq 0$ .
3. The problem is a linear programming problem.

Find the point on the line $x + 5y = 19$ when $x = 0$ .
$y = \frac{19}{5}$

Find the point on the line $x + 5y = 19$ when $y = 0$ .
$x = 19$

Find the point on the line $4x + y = 19$ when $x = 0$ .
$y = 19$

Find the point on the line $4x + y = 19$ when $y = 0$ .
$x = \frac{19}{4}$

Plot the lines and the feasible region on a coordinate plane. The feasible region is the area that satisfies all inequalities, including $x \geq 0$ and $y \geq 0$ .

Find the intersection point of the lines $x + 5y = 19$ and $4x + y = 19$ by solving the system of equations.

Set up the system of equations:
$\begin{align} x + 5y &= 19 \\ 4x + y &= 19 \end{align}$

Multiply the second equation by 5 to eliminate $y$ :
$\begin{align} x + 5y &= 19 \\ 20x + 5y &= 95 \end{align}$

Subtract the first equation from the second equation to solve for $x$ :
$\begin{align} (20x + 5y) - (x + 5y) &= 95 - 19 \\ 19x &= 76 \\ x &= 4 \end{align}$

Substitute $x = 4$ into the first equation to solve for $y$ :
$\begin{align} 4 + 5y &= 19 \\ 5y &= 15 \\ y &= 3 \end{align}$

The intersection point of the two lines is $(4, 3)$ .

Evaluate the objective function $c = x + 2y$ at each corner point of the feasible region, including $(4, 3)$ , $(19, 0)$ , $(0, \frac{19}{5})$ , and $(0, 19)$ .

Calculate $c$ at $(4, 3)$ :
$c = 4 + 2(3) = 4 + 6 = 10$

Calculate $c$ at $(19, 0)$ :
$c = 19 + 2(0) = 19$

Calculate $c$ at $(0, \frac{19}{5})$ :
$c = 0 + 2\left(\frac{19}{5}\right) = \frac{38}{5} = 7.6$

Calculate $c$ at $(0, 19)$ :
$c = 0 + 2(19) = 38$

The minimum value of $c$ is the smallest value obtained from the corner points.

Compare the values of $c$ to find the minimum:
- At $(4, 3)$ , $c = 10$ - At $(19, 0)$ , $c = 19$ - At $(0, \frac{19}{5})$ , $c = 7.6$ - At $(0, 19)$ , $c = 38$

The minimum value of $c$ is $7.6$ at the point $(0, \frac{19}{5})$ .
The solution to the problem is that the minimum value of the objective function $c$ is $7.6$ , and it occurs when $x = 0$ and $y = \frac{19}{5}$ .