Consider the LPP: Maximize $z = 5x + 3y$ subject to $3x + 5y \leq 15$, $5x + 2y \leq 10$, $x,y \geq 0$. The optimal feasible solution occurs at

The maximum value of $z = 5x + 3y$ occurs at a corner point of the feasible region formed by the constraints.

---

The corner points are found by checking the origin, axis intersections, and constraint line intersections.

At the origin: $(0, 0)$

---

For the intersection with the x-axis, set $y = 0$:

From $3x + 5y \leq 15$: $x \leq 5$

From $5x + 2y \leq 10$: $x \leq 2$

The binding constraint gives the corner point: $(2, 0)$

---

For the intersection with the y-axis, set $x = 0$:

From $3x + 5y \leq 15$: $y \leq 3$

From $5x + 2y \leq 10$: $y \leq 5$

The binding constraint gives the corner point: $(0, 3)$

---

For the intersection of the two constraint lines:

$3x + 5y = 15$ ... (1)

$5x + 2y = 10$ ... (2)

Multiply equation (1) by 2: $6x + 10y = 30$

Multiply equation (2) by 5: $25x + 10y = 50$

Subtracting: $19x = 20$

$x = \dfrac{20}{19}$

Substituting into equation (2):

$5 \cdot \dfrac{20}{19} + 2y = 10$

$\dfrac{100}{19} + 2y = 10$

$2y = \dfrac{90}{19}$

$y = \dfrac{45}{19}$

The corner point is: $\left(\dfrac{20}{19}, \dfrac{45}{19}\right)$

---

Evaluating $z = 5x + 3y$ at each corner point:

At $(0, 0)$: $z = 0$

At $(2, 0)$: $z = 10$

At $(0, 3)$: $z = 9$

At $\left(\dfrac{20}{19}, \dfrac{45}{19}\right)$:

$z = 5 \cdot \dfrac{20}{19} + 3 \cdot \dfrac{45}{19}$

$z = \dfrac{100 + 135}{19}$

$z = \dfrac{235}{19} \approx 12.37$

---

The maximum value is $\dfrac{235}{19}$, occurring at the point $\left(\dfrac{20}{19}, \dfrac{45}{19}\right)$.