Find all horizontal planes that are tangent to the surface z = x³ - 3xy² + 6y² + 1.

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Quizplus · Accepted Answer

To find the horizontal planes that are tangent to the given surface, we need to find the critical points of the function and then evaluate the function at those points. The given surface is $z = x^3 - 3xy^2 + 6y^2 + 1$. To find the critical points, we take the partial derivatives with respect to $x$ and $y$ and set them equal to zero.$\frac{\partial z}{\partial x} = 3x^2 - 3y^2$$\frac{\partial z}{\partial y} = -6xy + 12y$Setting these equal to zero gives:$3x^2 - 3y^2 = 0$ and $-6xy + 12y = 0$From the first equation, we get $x^2 = y^2$, which implies $x = y$ or $x = -y$ (since we're dealing with real numbers).Plugging $x = y$ into the second equation gives $-6y^2 + 12y = 0$, which simplifies to $y(6y - 12) = 0$, giving $y = 0$ or $y = 2$. Similarly, plugging $x = -y$ into the second equation gives the same result because the equation is symmetric in $x$ and $y$.Evaluating the function at these critical points:For $y = 0$, $x$ can be anything, but it doesn't affect the z-value since the equation simplifies to $z = x^3 + 1$, which does not give a constant $z$ for a horizontal plane.For $y = 2$, $x = 2$ or $x = -2$, plugging these into the original equation gives $z = 1$ and $z = 3$ respectively.Therefore, the horizontal planes that are tangent to the given surface are $z = 1$ and $z = 3$.

Find All Horizontal Planes That Are Tangent to the Surface