The maximum value of $$z$$ in the following equation $$z = 6xy + y^2$$, where $$3x + 4y \leq 100$$ and $$4x + 3y \leq 75$$ for $$x \geq 0$$ and $$y \geq 0$$ is ________.

We maximize $$z = 6xy + y^2$$ subject to $$3x + 4y \leq 100$$, $$4x + 3y \leq 75$$, $$x \geq 0$$, $$y \geq 0$$.

First we find the corner points of the feasible region. The constraints $$3x + 4y = 100$$ and $$4x + 3y = 75$$ intersect where: multiplying the first by 4 and the second by 3 gives $$12x + 16y = 400$$ and $$12x + 9y = 225$$. Subtracting: $$7y = 175$$, so $$y = 25$$ and $$x = 0$$. The feasible region has corners at $$(0,0)$$, $$(75/4, 0) = (18.75, 0)$$, and $$(0, 25)$$.

Evaluating $$z$$ at the corners: at $$(0,0)$$: $$z = 0$$. At $$(18.75, 0)$$: $$z = 0$$. At $$(0, 25)$$: $$z = 625$$.

Since $$z = 6xy + y^2$$ is not linear, we must also check the boundary edges. On the axes ($$x = 0$$ or $$y = 0$$), $$z$$ simplifies to $$y^2$$ or 0, so the maximum on these boundaries is at $$(0, 25)$$ with $$z = 625$$.

Now check the edge $$4x + 3y = 75$$ with $$0 \leq y \leq 25$$. Here $$x = \frac{75 - 3y}{4}$$. Substituting into $$z$$:

$$z = 6 \cdot \frac{75-3y}{4} \cdot y + y^2 = \frac{6y(75-3y)}{4} + y^2 = \frac{450y - 18y^2}{4} + y^2 = \frac{450y - 18y^2 + 4y^2}{4} = \frac{450y - 14y^2}{4}$$

Taking the derivative: $$\frac{dz}{dy} = \frac{450 - 28y}{4}$$. Setting this to zero: $$y = \frac{450}{28} = \frac{225}{14} \approx 16.07$$. The second derivative is $$\frac{-28}{4} = -7 < 0$$, confirming this is a maximum.

At $$y = \frac{225}{14}$$: $$x = \frac{75 - 675/14}{4} = \frac{375/14}{4} = \frac{375}{56} \approx 6.70$$. We verify $$3x + 4y = \frac{1125}{56} + \frac{900}{14} = \frac{1125 + 3600}{56} = \frac{4725}{56} \approx 84.4 \leq 100$$, so the point is feasible.

$$z = \frac{450 \cdot 225/14 - 14 \cdot (225/14)^2}{4} = \frac{101250/14 - 50625/14}{4} = \frac{50625}{56} \approx 904$$.

Comparing all candidates: corners give at most 625, while the boundary maximum gives $$\frac{50625}{56} \approx 904$$.

The maximum value of $$z$$ is 904.