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:

The correct option is 50.

We are given an objective function:

$$Z=ax+by$$

This function achieves its maximum value at two distinct points, $(8,2)$ and $(4,6)$. Since the maximum value of an objective function in a linear programming problem is unique, the value of $Z$ at both of these points must be equal.

Let us evaluate $Z$ at the point $(8,2)$:
$$Z_1=a\left(8\right)+b\left(2\right)=8a+2b$$

Next, let us evaluate $Z$ at the point $(4,6)$:
$$Z_2=a\left(4\right)+b\left(6\right)=4a+6b$$

Since the maximum value is the same at both points, we equate $Z_1$ and $Z_2$:
$$8a+2b=4a+6b$$

Subtracting $4a$ and $2b$ from both sides, we get:
$$4a=4b$$

Dividing by 4, we find:
$$a=b$$

We are given that $a\ge 0$, $b\ge 0$, and the product:
$$ab=25$$

Substituting $b=a$ into the product equation gives:
$$a^2=25$$

Taking the positive square root (since $a\ge 0$):
$$a=5$$

Since $a=b$, we also have $b=5$.

Now, we calculate the maximum value of $Z$ by substituting these values of $a$ and $b$ back into the expression for $Z$ at one of the points, say $(8,2)$:
$$Z_{\text{max}}=8a+2b=8\left(5\right)+2\left(5\right)=40+10=50$$

Thus, the maximum value of the objective function is indeed 50.