Answer:
We have the coldest value of temperature [tex]T(\frac{3}{4},0) = -9/16 [/tex]. and the hottest value is [tex]T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}[/tex].
Step-by-step explanation:
We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.
The temperature equation is:
[tex]T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x[/tex]
Let's take the partials derivatives.
[tex]T_{x}(x,y)=2x-\frac{3}{2}=0[/tex]
[tex]T_{y}(x,y)=4y=0[/tex]
So, we can find the critical point (x,y) of T(x,y).
[tex]2x-\frac{3}{2}=0[/tex]
[tex]x=\frac{3}{4}[/tex]
[tex]4y=0[/tex]
[tex]y=0[/tex]
The critical point is (3/4,0) so the temperature at this point is: [tex]T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})[/tex]
[tex]T(\frac{3}{4},0)=-\frac{9}{16}[/tex]
Now, we need to evaluate the boundary condition.
[tex]x^{2}+y^{2}=1[/tex]
We can solve this equation for y and evaluate this value in the temperature.
[tex]y=\pm \sqrt{1-x^{2}}[/tex]
[tex]T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x[/tex]
[tex]T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2[/tex]
Now, let's find the critical point again, as we did above.
[tex]T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0[/tex]
[tex]x=-\frac{3}{4}[/tex]
Evaluating T(x,y) at this point, we have:
[tex]T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2[/tex]
[tex]T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}[/tex]
Now, we can see that at point (3/4,0) we have the coldest value of temperature [tex]T(\frac{3}{4},0) = -9/16 [/tex]. On the other hand, at the point [tex]-(3/4),\frac{\sqrt{7}}{4})[/tex] we have the hottest value of temperature, it is [tex]T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}[/tex].
I hope it helps you!
