Solved on Dec 12, 2023

Maximize $9t + 8.50v$ under constraints $9t + 8.50v \leq 150$ and $v \geq 5$ .

STEP 1

Assumptions
1. We have a system of inequalities with two variables $t$ and $v$ .
2. The first inequality is $9t + 8.50v \leq 150$ .
3. The second inequality is $v \geq 5$ .
4. We need to find the possible values of $t$ and $v$ that satisfy both inequalities.

STEP 2

We will first focus on the second inequality, which is simpler.
$v \geq 5$
This inequality tells us that the value of $v$ must be greater than or equal to 5.

STEP 3

Now, we will manipulate the first inequality to express $t$ in terms of $v$ .
$9t + 8.50v \leq 150$
Subtract $8.50v$ from both sides of the inequality to isolate terms involving $t$ .
$9t \leq 150 - 8.50v$

STEP 4

Next, divide both sides of the inequality by 9 to solve for $t$ .
$t \leq \frac{150 - 8.50v}{9}$

STEP 5

Since we know that $v \geq 5$ , we can substitute $v = 5$ into the inequality to find the maximum value of $t$ when $v$ is at its minimum.
$t \leq \frac{150 - 8.50(5)}{9}$

STEP 6

Calculate the value inside the parentheses.
$8.50(5) = 42.50$

STEP 7

Substitute this value back into the inequality.
$t \leq \frac{150 - 42.50}{9}$

STEP 8

Perform the subtraction in the numerator.
$t \leq \frac{107.50}{9}$

STEP 9

Finally, divide to find the maximum value of $t$ when $v$ is 5.
$t \leq \frac{107.50}{9} \approx 11.94$

STEP 10

Now we have found that when $v$ is at its minimum value of 5, $t$ can be at most approximately 11.94. However, as $v$ increases, the maximum value of $t$ will decrease, since $t$ and $v$ are inversely related in the first inequality.

STEP 11

To find the range of possible values for $t$ and $v$ , we can graph the inequalities on a coordinate plane where $t$ is on the x-axis and $v$ is on the y-axis. The solution will be the shaded region that satisfies both inequalities.

STEP 12

The line $9t + 8.50v = 150$ is the boundary for the first inequality. If we graph this line, any point on or below this line will satisfy the first inequality.

STEP 13

The line $v = 5$ is the boundary for the second inequality. If we graph this line, any point above this line will satisfy the second inequality.

STEP 14

The intersection of the regions defined by these two inequalities will give us the set of all possible solutions for $t$ and $v$ .
To summarize, the solution to the system of inequalities is the set of all points $(t, v)$ such that $t \leq \frac{150 - 8.50v}{9}$ and $v \geq 5$ , which can be represented graphically as the shaded region in the coordinate plane where the first inequality's region is below the line $9t + 8.50v = 150$ and the second inequality's region is above the line $v = 5$ .

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Maximize $9t + 8.50v$ under constraints $9t + 8.50v \leq 150$ and $v \geq 5$ .

STEP 1

Assumptions
1. We have a system of inequalities with two variables $t$ and $v$ .
2. The first inequality is $9t + 8.50v \leq 150$ .
3. The second inequality is $v \geq 5$ .
4. We need to find the possible values of $t$ and $v$ that satisfy both inequalities.

STEP 2

We will first focus on the second inequality, which is simpler.
$v \geq 5$
This inequality tells us that the value of $v$ must be greater than or equal to 5.

STEP 3

Now, we will manipulate the first inequality to express $t$ in terms of $v$ .
$9t + 8.50v \leq 150$
Subtract $8.50v$ from both sides of the inequality to isolate terms involving $t$ .
$9t \leq 150 - 8.50v$

STEP 4

Next, divide both sides of the inequality by 9 to solve for $t$ .
$t \leq \frac{150 - 8.50v}{9}$

STEP 5

Since we know that $v \geq 5$ , we can substitute $v = 5$ into the inequality to find the maximum value of $t$ when $v$ is at its minimum.
$t \leq \frac{150 - 8.50(5)}{9}$

STEP 6

Calculate the value inside the parentheses.
$8.50(5) = 42.50$

STEP 7

Substitute this value back into the inequality.
$t \leq \frac{150 - 42.50}{9}$

STEP 8

Perform the subtraction in the numerator.
$t \leq \frac{107.50}{9}$

STEP 9

Finally, divide to find the maximum value of $t$ when $v$ is 5.
$t \leq \frac{107.50}{9} \approx 11.94$

STEP 10

Now we have found that when $v$ is at its minimum value of 5, $t$ can be at most approximately 11.94. However, as $v$ increases, the maximum value of $t$ will decrease, since $t$ and $v$ are inversely related in the first inequality.

STEP 11

To find the range of possible values for $t$ and $v$ , we can graph the inequalities on a coordinate plane where $t$ is on the x-axis and $v$ is on the y-axis. The solution will be the shaded region that satisfies both inequalities.

STEP 12

The line $9t + 8.50v = 150$ is the boundary for the first inequality. If we graph this line, any point on or below this line will satisfy the first inequality.

STEP 13

The line $v = 5$ is the boundary for the second inequality. If we graph this line, any point above this line will satisfy the second inequality.

STEP 14

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Maximize 9t+8.50v9t + 8.50v9t+8.50v under constraints 9t+8.50v≤1509t + 8.50v \leq 1509t+8.50v≤150 and v≥5v \geq 5v≥5.

STEP 1

Assumptions1. We have a system of inequalities with two variables ttt and vvv.2. The first inequality is 9t+8.50v≤1509t + 8.50v \leq 1509t+8.50v≤150.3. The second inequality is v≥5v \geq 5v≥5.4. We need to find the possible values of ttt and vvv that satisfy both inequalities.

STEP 2

We will first focus on the second inequality, which is simpler.v≥5v \geq 5v≥5This inequality tells us that the value of vvv must be greater than or equal to 5.

STEP 3

Now, we will manipulate the first inequality to express ttt in terms of vvv.9t+8.50v≤1509t + 8.50v \leq 1509t+8.50v≤150Subtract 8.50v8.50v8.50v from both sides of the inequality to isolate terms involving ttt.9t≤150−8.50v9t \leq 150 - 8.50v9t≤150−8.50v

STEP 4

Next, divide both sides of the inequality by 9 to solve for ttt.t≤150−8.50v9t \leq \frac{150 - 8.50v}{9}t≤9150−8.50v​

STEP 5

Since we know that v≥5v \geq 5v≥5, we can substitute v=5v = 5v=5 into the inequality to find the maximum value of ttt when vvv is at its minimum.t≤150−8.50(5)9t \leq \frac{150 - 8.50(5)}{9}t≤9150−8.50(5)​

STEP 6

Calculate the value inside the parentheses.8.50(5)=42.508.50(5) = 42.508.50(5)=42.50

STEP 7

Substitute this value back into the inequality.t≤150−42.509t \leq \frac{150 - 42.50}{9}t≤9150−42.50​

STEP 8

Perform the subtraction in the numerator.t≤107.509t \leq \frac{107.50}{9}t≤9107.50​

STEP 9

Finally, divide to find the maximum value of ttt when vvv is 5.t≤107.509≈11.94t \leq \frac{107.50}{9} \approx 11.94t≤9107.50​≈11.94

STEP 10

Now we have found that when vvv is at its minimum value of 5, ttt can be at most approximately 11.94. However, as vvv increases, the maximum value of ttt will decrease, since ttt and vvv are inversely related in the first inequality.

STEP 11

To find the range of possible values for ttt and vvv, we can graph the inequalities on a coordinate plane where ttt is on the x-axis and vvv is on the y-axis. The solution will be the shaded region that satisfies both inequalities.

STEP 12

The line 9t+8.50v=1509t + 8.50v = 1509t+8.50v=150 is the boundary for the first inequality. If we graph this line, any point on or below this line will satisfy the first inequality.

STEP 13

The line v=5v = 5v=5 is the boundary for the second inequality. If we graph this line, any point above this line will satisfy the second inequality.

STEP 14

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Maximize $9t + 8.50v$ under constraints $9t + 8.50v \leq 150$ and $v \geq 5$ .

Assumptions
1. We have a system of inequalities with two variables $t$ and $v$ .
2. The first inequality is $9t + 8.50v \leq 150$ .
3. The second inequality is $v \geq 5$ .
4. We need to find the possible values of $t$ and $v$ that satisfy both inequalities.

We will first focus on the second inequality, which is simpler.
$v \geq 5$
This inequality tells us that the value of $v$ must be greater than or equal to 5.

Now, we will manipulate the first inequality to express $t$ in terms of $v$ .
$9t + 8.50v \leq 150$
Subtract $8.50v$ from both sides of the inequality to isolate terms involving $t$ .
$9t \leq 150 - 8.50v$

Next, divide both sides of the inequality by 9 to solve for $t$ .
$t \leq \frac{150 - 8.50v}{9}$

Since we know that $v \geq 5$ , we can substitute $v = 5$ into the inequality to find the maximum value of $t$ when $v$ is at its minimum.
$t \leq \frac{150 - 8.50(5)}{9}$

Calculate the value inside the parentheses.
$8.50(5) = 42.50$

Substitute this value back into the inequality.
$t \leq \frac{150 - 42.50}{9}$

Perform the subtraction in the numerator.
$t \leq \frac{107.50}{9}$

Finally, divide to find the maximum value of $t$ when $v$ is 5.
$t \leq \frac{107.50}{9} \approx 11.94$

Now we have found that when $v$ is at its minimum value of 5, $t$ can be at most approximately 11.94. However, as $v$ increases, the maximum value of $t$ will decrease, since $t$ and $v$ are inversely related in the first inequality.

To find the range of possible values for $t$ and $v$ , we can graph the inequalities on a coordinate plane where $t$ is on the x-axis and $v$ is on the y-axis. The solution will be the shaded region that satisfies both inequalities.

The line $9t + 8.50v = 150$ is the boundary for the first inequality. If we graph this line, any point on or below this line will satisfy the first inequality.

The line $v = 5$ is the boundary for the second inequality. If we graph this line, any point above this line will satisfy the second inequality.