The point which provides the optimal solution of the linear programming problem

maximize $z = 21x + 35y$

$3x + 2y \leq 30$

$4x + 5y \leq 60$

$x \geq 0, y \geq 0$

has the coordinates

In linear programming, the maximum or minimum value occurs at a corner point (vertex) of the feasible region.

The corner points are formed by the intersection of constraint lines and axes.

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At the origin:

Point: $(0, 0)$

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Where the first constraint meets the x-axis, set $y = 0$ in $3x + 2y = 30$:

$3x = 30$

$x = 10$

Point: $(10, 0)$

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Where the second constraint meets the y-axis, set $x = 0$ in $4x + 5y = 60$:

$5y = 60$

$y = 12$

Point: $(0, 12)$

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Where both constraints intersect, solve:

$3x + 2y = 30$ ... (1)

$4x + 5y = 60$ ... (2)

Multiply (1) by 5:

$15x + 10y = 150$

Multiply (2) by 2:

$8x + 10y = 120$

Subtract:

$7x = 30$

$x = \dfrac{30}{7}$

Substitute into (1):

$3\left(\dfrac{30}{7}\right) + 2y = 30$

$2y = \dfrac{120}{7}$

$y = \dfrac{60}{7}$

Point: $\left(\dfrac{30}{7}, \dfrac{60}{7}\right)$

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Evaluate $z = 21x + 35y$ at each corner point:

At $(0, 0)$: $z = 21(0) + 35(0) = 0$

At $(10, 0)$: $z = 21(10) + 35(0) = 210$

At $(0, 12)$: $z = 21(0) + 35(12) = 420$

At $\left(\dfrac{30}{7}, \dfrac{60}{7}\right)$: $z = 21\left(\dfrac{30}{7}\right) + 35\left(\dfrac{60}{7}\right) = \dfrac{630}{7} + \dfrac{2100}{7} = \dfrac{2730}{7} = 390$

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The maximum value of $z = 420$ occurs at point $(0, 12)$.

Therefore, the coordinates are $(0, 12)$.