Question

Temperature of an elliptical plate The temperature of points on an elliptical plate $x^2+y^2+xy\le 1$ is given by $T(x,y)=25(x^2+y^2).$ Find the hottest and coldest temperatures on the edge of the plate.

Short answer

Answer: The hottest and coldest temperatures on the edge of the elliptical plate are both $50/3$ degrees.

Step-by-step solution

Find the equation of the ellipse representing the edge of the plate

The equation of the ellipse is given by $x^2+y^2+xy=1$. This represents the edge of the elliptical plate.

Set up the Lagrange Multiplier equation for the function and constraint

We want to optimize the temperature function $T(x,y)=25(x^2+y^2)$, subject to the constraint $x^2+y^2+xy=1$. We introduce a Lagrange Multiplier $\lambda$ and form the function: $F(x,y,\lambda )=25(x^2+y^2)-\lambda (x^2+y^2+xy-1)$.

Solve the system of partial derivatives

To find the critical points, we need to take the partial derivatives of $F$ with respect to $x$, $y$, and $\lambda$, and set them equal to zero: 1) With respect to x: $\frac{\partial F}{\partial x}=50x-2\lambda x-\lambda y=0$. 2) With respect to y: $\frac{\partial F}{\partial y}=50y-2\lambda y-\lambda x=0$. 3) With respect to $\lambda$: $\frac{\partial F}{\partial \lambda }=x^2+y^2+xy-1=0$. Now, solve the system of equations (1), (2), and (3).

Solve for x, y, and λ

We can solve the system of equations in step 3 by substitution or elimination methods. For brevity, we provide the solutions: 1) $\lambda =50$ 2) $x=\pm \frac{1}{\sqrt{3}}$, $y=\mp \frac{1}{\sqrt{3}}$. These solutions correspond to the points at which the temperature reaches its maximum and minimum.

Find the temperatures at the critical points

Now, we can substitute the values of $x$ and $y$ back into the function $T(x,y)=25(x^2+y^2)$ to find the hottest and coldest temperatures: 1) Hottest temperature: $T(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})=25(2\cdot \frac{1}{3})=50/3$. 2) Coldest temperature: $T(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})=25(2\cdot \frac{1}{3})=50/3$. Hence, the hottest and coldest temperature on the edge of the elliptical plate are both $50/3$ degrees.

Key concepts

Optimization

Optimization is a fundamental concept in calculus, where we try to find the maximum or minimum value of a function within a given set of constraints. In this exercise, we want to find the hottest and coldest temperatures on the edge of an elliptical plate. This is an example of constrained optimization, where the temperature function, given by $T(x,y)=25(x^2+y^2)$, needs to be maximized or minimized while staying on the edge defined by the ellipse equation $x^2+y^2+xy=1$.
The method of Lagrange multipliers is a powerful technique used here to solve the optimization problem. It allows us to find the extremum of a function subject to constraints by adding an extra variable, the Lagrange multiplier $\lambda$. We form a new function $F(x,y,\lambda )=25(x^2+y^2)-\lambda (x^2+y^2+xy-1)$.

• Taking partial derivatives of $F$ with respect to $x$, $y$, and $\lambda$ gives us a system of equations.
• Solving these equations leads to critical points, which provide potential solutions for the hottest and coldest temperatures.

Thus, optimization using Lagrange multipliers elegantly provides the solution by handling constraints effectively.

Elliptical Plate

An elliptical plate in mathematical terms is a two-dimensional set bounded by an ellipse. The equation $x^2+y^2+xy=1$ describes the boundary of this plate. Ellipses are curves that encapsulate conic sections, and unlike circles, they have two axes of symmetry—the major and minor axes.
Understanding the equation of the ellipse helps in visualizing the boundary where the temperature extremes need to be found in this problem. Here, the elliptical plate equation sets restrictions on the values $x$ and $y$ can take. It essentially tells us that our points of interest lie exactly on this boundary, not inside or outside.

• Simultaneous equations derived from the system respect this boundary, ensuring that solutions are viable in the defined elliptical constraint.
• The points found through solving these are where the temperature (determined by another function) is to be evaluated.

This gives us a neat geometric interpretation intertwined with algebraic problem-solving for optimization on an ellipse.

Temperature

Temperature is a measure of heat intensity and is expressed here by the function $T(x,y)=25(x^2+y^2)$. This quadratic function models how temperature varies across the elliptical plate.
The temperature function assigns a temperature to every point $(x,y)$ on the elliptical plate. In this exercise, we focus on finding extreme temperature values on the edge. The key challenge is balancing how temperature peaks and troughs interact with the constraint that is the edge of the ellipse.

• Once we know the edge equation, we spot where temperature reaches maximum or minimum values.
• The method adopted here models how temperature relies solely on spatial data (the geometric parameters $x$ and $y$).

Both the hottest and coldest points have been calculated analytically, showing the symmetry here in terms of heat distribution over the elliptical boundary. This displays how straightforward calculus can provide insights into seemingly complex thermal behaviors.