An objective function Z = ax + by is maximum at points (8, 2) and (4, 6). If a ≥ 0, b ≥ 0, and ab = 25, then the maximum value of the function is:
50
The question involves a concept from Linear Programming, specifically dealing with an objective function $Z = ax + by$ that is being maximized. We are given that the maximum value of this function occurs at two distinct points, (8, 2) and (4, 6), within the feasible region. We are also given constraints on the coefficients $a$ and $b$: $a \ge 0$, $b \ge 0$, and $ab = 25$. Our goal is to find the maximum value of the function $Z$.
In Linear Programming, if an objective function attains its maximum (or minimum) value at two distinct corner points of the feasible region, then it attains the same maximum (or minimum) value at every point on the line segment connecting these two corner points. This occurs when the objective function line is parallel to the line segment connecting the two points.
This means the slope of the objective function line $ax + by = Z_{max}$ must be equal to the slope of the line segment connecting the points (8, 2) and (4, 6).
The slope of a line segment connecting two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$\text{Slope} = \frac{y_2 - y_1}{x_2 - x_1}$
Using the points (8, 2) and (4, 6):
$(x_1, y_1) = (8, 2)$
$(x_2, y_2) = (4, 6)$
$\text{Slope of segment} = \frac{6 - 2}{4 - 8} = \frac{4}{-4} = -1$
The objective function is $Z = ax + by$. To find the slope of the objective function line $ax + by = Z$, we can rearrange it into slope-intercept form ($y = mx + c$):
$by = -ax + Z$
$y = -\frac{a}{b}x + \frac{Z}{b}$
The slope of the objective function line is $-\frac{a}{b}$.
Since the objective function is maximum at all points on the segment connecting (8, 2) and (4, 6), its slope must equal the slope of this segment:
$-\frac{a}{b} = -1$
This simplifies to:
$\frac{a}{b} = 1$
$a = b$
We are given the condition $ab = 25$. We have also found that $a = b$. Substitute $a$ for $b$ in the equation $ab = 25$:
$a \cdot a = 25$
$a^2 = 25$
Taking the square root of both sides:
$a = \pm \sqrt{25}$
$a = \pm 5$
We are given the constraint $a \ge 0$. Therefore, we must choose the positive value:
$a = 5$
Since $a = b$, we also have:
$b = 5$
These values ($a=5, b=5$) satisfy the conditions $a \ge 0$, $b \ge 0$, and $ab = 5 \times 5 = 25$.
Now that we have the values of $a$ and $b$, the objective function is $Z = 5x + 5y$.
To find the maximum value, we can evaluate $Z$ at either of the given points where the maximum occurs, (8, 2) or (4, 6).
Evaluating at point (8, 2):
$Z = 5(8) + 5(2)$
$Z = 40 + 10$
$Z = 50$
Evaluating at point (4, 6):
$Z = 5(4) + 5(6)$
$Z = 20 + 30$
$Z = 50$
Both points yield the same maximum value.
| Point (x, y) | Objective Function $Z = 5x + 5y$ | Value of Z |
|---|---|---|
| (8, 2) | $5(8) + 5(2)$ | 50 |
| (4, 6) | $5(4) + 5(6)$ | 50 |
The maximum value of the function is 50.
| Step | Description | Action Taken |
|---|---|---|
| 1 | Identify given information | Objective function $Z=ax+by$, max at (8,2) and (4,6), $a \ge 0, b \ge 0, ab=25$. |
| 2 | Understand implication of max at two points | Slope of objective function line equals slope of line segment connecting the points. |
| 3 | Calculate slope of segment | $\frac{6-2}{4-8} = -1$. |
| 4 | Relate objective function slope to segment slope | $-\frac{a}{b} = -1 \implies a = b$. |
| 5 | Use constraint to find a and b | $ab=25$ and $a=b \implies a^2=25$. With $a \ge 0$, $a=5$, so $b=5$. |
| 6 | Formulate specific objective function | $Z = 5x + 5y$. |
| 7 | Calculate max value | Evaluate $Z$ at (8, 2) or (4, 6). $Z = 5(8) + 5(2) = 50$. |
Linear Programming (LP) is a mathematical method for determining a way to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships. Key components include:
This problem highlights the property that if optimality occurs at two vertices, the objective function's contour line at the optimal value is parallel to the edge connecting these vertices.
The corner points of the feasible region determined by:
x + y ≤ 8, 2x + y ≥ 8, x ≥ 0, y ≥ 0
are A(0, 8), B(4, 0), and C(8, 0). If the objective function Z = ax + by has its maximum value on the line segment AB, then the relation between a and b is:
Which of the following cannot be the direction ratios of the straight line:
\( \frac{x - 3}{2} = \frac{2 - y}{3} = \frac{z + 4}{-1} \)
Corner points of a feasible bounded region are (0,10), (4,2), (3,7) and (10,6). Maximum value 50 of objective function \( z = ax + by \) occurs at two points (0,10) and (10,6). The value of \( a \) and \( b \) are:
The corner points of the feasible region determined by system of linear constraints are (60,0), (120,0), (40,20), and (60,30). Let \( z = ax + by \), where \( a, b > 0 \) be the objective function. Find condition on \( a \) and \( b \) so that the maximum of \( z \) occurs at (120,0) and (60,30).
The feasible region corresponding to an LPP represented by the constraints \( x \geq 7, y \geq 4, x + 2y \geq 8 \) is: