1. **State the problem:** Find the stationary points of the function $$f(x,y) = x^3 + 3xy^2 - 15x^2 - 15y^2 + 72x$$. 2. **Formula and rules:** Stationary points occur where the gradient (partial derivatives) equals zero: $$\frac{\partial f}{\partial x} = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 0$$ 3. **Calculate partial derivatives:** $$\frac{\partial f}{\partial x} = 3x^2 + 3y^2 - 30x + 72$$ $$\frac{\partial f}{\partial y} = 6xy - 30y$$ 4. **Set partial derivatives to zero:** $$3x^2 + 3y^2 - 30x + 72 = 0$$ $$6xy - 30y = 0$$ 5. **Solve the second equation:** $$6y(x - 5) = 0 \implies y=0 \quad \text{or} \quad x=5$$ 6. **Case 1: $y=0$** Substitute into first equation: $$3x^2 - 30x + 72 = 0$$ Divide by 3: $$x^2 - 10x + 24 = 0$$ Factor: $$(x - 6)(x - 4) = 0$$ So, $$x=6$$ or $$x=4$$ Stationary points: $$(6,0), (4,0)$$ 7. **Case 2: $x=5$** Substitute into first equation: $$3(5)^2 + 3y^2 - 30(5) + 72 = 0$$ Calculate: $$75 + 3y^2 - 150 + 72 = 0$$ Simplify: $$3y^2 - 3 = 0$$ $$3y^2 = 3$$ $$y^2 = 1$$ $$y = \pm 1$$ Stationary points: $$(5,1), (5,-1)$$ 8. **Summary:** Stationary points are $$(6,0), (4,0), (5,1), (5,-1)$$. 9. **Check given options:** The only points matching the options are $$(6,0)$$ and $$(4,0)$$. **Final answer:** $$(4,0), (6,0)$$