1. **State the problem:** We want to maximize the function $$f(x,y) = 5y^2 - xy$$ subject to the constraint $$x = y^3 - y^2$$ and $$y \geq -2$$. 2. **Substitute the constraint into the function:** Replace $$x$$ in $$f(x,y)$$ with $$y^3 - y^2$$ to get a function of $$y$$ only: $$f(y) = 5y^2 - (y^3 - y^2) y = 5y^2 - y(y^3 - y^2) = 5y^2 - (y^4 - y^3) = 5y^2 - y^4 + y^3$$ 3. **Simplify the function:** $$f(y) = -y^4 + y^3 + 5y^2$$ 4. **Find critical points by differentiating:** $$f'(y) = \frac{d}{dy}(-y^4 + y^3 + 5y^2) = -4y^3 + 3y^2 + 10y$$ 5. **Set derivative equal to zero to find critical points:** $$-4y^3 + 3y^2 + 10y = 0$$ Factor out $$y$$: $$y(-4y^2 + 3y + 10) = 0$$ So either $$y=0$$ or solve quadratic: $$-4y^2 + 3y + 10 = 0$$ Multiply both sides by $$\cancel{-1}$$: $$4y^2 - 3y - 10 = 0$$ 6. **Solve quadratic equation:** Using quadratic formula: $$y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-10)}}{2 \cdot 4} = \frac{3 \pm \sqrt{9 + 160}}{8} = \frac{3 \pm \sqrt{169}}{8} = \frac{3 \pm 13}{8}$$ 7. **Calculate roots:** $$y_1 = \frac{3 + 13}{8} = \frac{16}{8} = 2$$ $$y_2 = \frac{3 - 13}{8} = \frac{-10}{8} = -\frac{5}{4} = -1.25$$ 8. **List critical points:** $$y = 0, y = 2, y = -1.25$$ 9. **Check domain constraint:** $$y \geq -2$$, all critical points satisfy this. 10. **Evaluate $$f(y)$$ at critical points:** - At $$y=0$$: $$f(0) = -0 + 0 + 0 = 0$$ - At $$y=2$$: $$f(2) = -(2)^4 + (2)^3 + 5(2)^2 = -16 + 8 + 20 = 12$$ - At $$y=-1.25$$: $$f(-1.25) = -(-1.25)^4 + (-1.25)^3 + 5(-1.25)^2 = -2.4414 -1.9531 + 7.8125 = 3.418$$ (approximate) 11. **Check endpoint $$y = -2$$:** $$f(-2) = -(-2)^4 + (-2)^3 + 5(-2)^2 = -16 -8 + 20 = -4$$ 12. **Determine maximum value:** The maximum value is $$12$$ at $$y=2$$. 13. **Find corresponding $$x$$:** $$x = y^3 - y^2 = 2^3 - 2^2 = 8 - 4 = 4$$ **Final answer:** The maximum value is **12**. It occurs when $$x = 4$$ and $$y = 2$$.