1. **State the problem:** We are given the surface equation $$z = -2y^2 - x^2 + 54$$ and boundary planes $$x = 3$$ and $$y = 3$$. We want to understand the shape and behavior of this paraboloid within these boundaries. 2. **Identify the surface type:** The equation is a quadratic form in $$x$$ and $$y$$ with negative coefficients, indicating a downward-opening paraboloid. 3. **Analyze cross sections:** - For the plane $$x = 3$$, substitute $$x = 3$$ into the equation: $$z = -2y^2 - (3)^2 + 54 = -2y^2 - 9 + 54 = -2y^2 + 45$$ This is a parabola opening downward in $$y$$. - For the plane $$y = 3$$, substitute $$y = 3$$: $$z = -2(3)^2 - x^2 + 54 = -18 - x^2 + 54 = -x^2 + 36$$ This is a parabola opening downward in $$x$$. 4. **Find vertex:** The vertex of the paraboloid is at $$x=0, y=0$$, where $$z = -2(0)^2 - 0^2 + 54 = 54$$ This is the maximum point. 5. **Summary:** The paraboloid opens downward with maximum height 54 at the origin. The boundary planes $$x=3$$ and $$y=3$$ restrict the domain, so the surface is considered only where $$x \leq 3$$ and $$y \leq 3$$. Final answer: The surface is a downward-opening paraboloid $$z = -2y^2 - x^2 + 54$$ with maximum at $$(0,0,54)$$, bounded by planes $$x=3$$ and $$y=3$$.