Math / Algebra

QuestionThe comer points for the bounded feasible region determined by the system of inequalities shown below are $O=(0,0), A=(0,7), B=(2,6)$ , and $C=(4,0)$ . $\begin{aligned} x+2 y & \leq 14 \\ 3 x+y & \leq 12 \\ x, y & \geq 0 \end{aligned}$
If $P=a x+b y$ and $a, b>0$ , determine conditions on $a$ and $b$ that will ensure that the maximum value of $P$ occurs only at $B$ .
Which conditions on a and b will ensure that the maximum value of $P$ occurs at B? A. $a>3 b$ B. $\frac{b}{2}<a<3 b$ C. $b=\frac{a}{2}$ D. $b>3 a$ E. $a=3 b$ F. $\frac{a}{2}<b<3 a$ G. $b<\frac{a}{2}$ H. $a<\frac{b}{2}$
1. $\mathrm{b}=3 \mathrm{a}$ J. $a=\frac{b}{2}$

Studdy Solution

STEP 1

1. The feasible region is determined by the given inequalities.
2. The objective function is $P = ax + by$ with $a, b > 0$ .
3. We need to find conditions on $a$ and $b$ such that the maximum value of $P$ occurs only at point $B = (2, 6)$ .

STEP 2

1. Determine the value of $P$ at each corner point.
2. Set up inequalities based on the condition that $P$ is maximized only at $B$ .
3. Solve the inequalities to find the conditions on $a$ and $b$ .

STEP 3

Calculate the value of $P$ at each corner point: - At $O = (0, 0)$ , $P = a \cdot 0 + b \cdot 0 = 0$ . - At $A = (0, 7)$ , $P = a \cdot 0 + b \cdot 7 = 7b$ . - At $B = (2, 6)$ , $P = a \cdot 2 + b \cdot 6 = 2a + 6b$ . - At $C = (4, 0)$ , $P = a \cdot 4 + b \cdot 0 = 4a$ .

STEP 4

Set up inequalities to ensure $P$ is maximized only at $B$ :
1. $2a + 6b > 7b$ (so $B$ is better than $A$ )
2. $2a + 6b > 4a$ (so $B$ is better than $C$ )

STEP 5

Solve the inequalities:
1. From $2a + 6b > 7b$ , we get $2a > b$ or $a > \frac{b}{2}$ .
2. From $2a + 6b > 4a$ , we get $6b > 2a$ or $b > \frac{a}{3}$ .

STEP 6

Combine the conditions $a > \frac{b}{2}$ and $b > \frac{a}{3}$ to find the correct option: - The condition $\frac{b}{2} < a < 3b$ satisfies both $a > \frac{b}{2}$ and $b > \frac{a}{3}$ .
The conditions on $a$ and $b$ that ensure the maximum value of $P$ occurs only at $B$ are:
$\boxed{\frac{b}{2} < a < 3b}$

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Studdy Solution

STEP 1

1. The feasible region is determined by the given inequalities.
2. The objective function is $P = ax + by$ with $a, b > 0$ .
3. We need to find conditions on $a$ and $b$ such that the maximum value of $P$ occurs only at point $B = (2, 6)$ .

STEP 2

1. Determine the value of $P$ at each corner point.
2. Set up inequalities based on the condition that $P$ is maximized only at $B$ .
3. Solve the inequalities to find the conditions on $a$ and $b$ .

STEP 3

Calculate the value of $P$ at each corner point: - At $O = (0, 0)$ , $P = a \cdot 0 + b \cdot 0 = 0$ . - At $A = (0, 7)$ , $P = a \cdot 0 + b \cdot 7 = 7b$ . - At $B = (2, 6)$ , $P = a \cdot 2 + b \cdot 6 = 2a + 6b$ . - At $C = (4, 0)$ , $P = a \cdot 4 + b \cdot 0 = 4a$ .

STEP 4

Set up inequalities to ensure $P$ is maximized only at $B$ :
1. $2a + 6b > 7b$ (so $B$ is better than $A$ )
2. $2a + 6b > 4a$ (so $B$ is better than $C$ )

STEP 5

Solve the inequalities:
1. From $2a + 6b > 7b$ , we get $2a > b$ or $a > \frac{b}{2}$ .
2. From $2a + 6b > 4a$ , we get $6b > 2a$ or $b > \frac{a}{3}$ .

STEP 6

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Studdy Solution

STEP 1

1. The feasible region is determined by the given inequalities.2. The objective function is P=ax+by P = ax + by P=ax+by with a,b>0 a, b > 0 a,b>0.3. We need to find conditions on a a a and b b b such that the maximum value of P P P occurs only at point B=(2,6) B = (2, 6) B=(2,6).

STEP 2

1. Determine the value of P P P at each corner point.2. Set up inequalities based on the condition that P P P is maximized only at B B B.3. Solve the inequalities to find the conditions on a a a and b b b.

STEP 3

STEP 4

Set up inequalities to ensure P P P is maximized only at B B B:1. 2a+6b>7b 2a + 6b > 7b 2a+6b>7b (so B B B is better than A A A)2. 2a+6b>4a 2a + 6b > 4a 2a+6b>4a (so B B B is better than C C C)

STEP 5

Solve the inequalities:1. From 2a+6b>7b 2a + 6b > 7b 2a+6b>7b, we get 2a>b 2a > b 2a>b or a>b2 a > \frac{b}{2} a>2b​.2. From 2a+6b>4a 2a + 6b > 4a 2a+6b>4a, we get 6b>2a 6b > 2a 6b>2a or b>a3 b > \frac{a}{3} b>3a​.

STEP 6

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The feasible region is determined by the given inequalities.
2. The objective function is $P = ax + by$ with $a, b > 0$ .
3. We need to find conditions on $a$ and $b$ such that the maximum value of $P$ occurs only at point $B = (2, 6)$ .

1. Determine the value of $P$ at each corner point.
2. Set up inequalities based on the condition that $P$ is maximized only at $B$ .
3. Solve the inequalities to find the conditions on $a$ and $b$ .

Set up inequalities to ensure $P$ is maximized only at $B$ :
1. $2a + 6b > 7b$ (so $B$ is better than $A$ )
2. $2a + 6b > 4a$ (so $B$ is better than $C$ )

Solve the inequalities:
1. From $2a + 6b > 7b$ , we get $2a > b$ or $a > \frac{b}{2}$ .
2. From $2a + 6b > 4a$ , we get $6b > 2a$ or $b > \frac{a}{3}$ .