Here is one way.

I don't know how your lecturer explained the topic and how to solve problems regarding them. He might be using partial derivatives or whatever.

This way here is like treating the multi-variable function the way I approach single-variable functions. It might be different from your lecturer's approach.

T(x,y)= 36 - (1/5)[x^2 + (y - 5)^2]

i)What is the domain of the temperature function?

The domain is what defines the function.

What values of x and y in the 10 by 10 room that will give a real T(x,y)?

x = 0 to 10, and y = 0 to 10 also

So, domain is all values of x, and/or y, from 0 to 10 ft.

x = [0,10] ft.

y = [0,10] ft.

ii)Where is the likely location of the heater?

T(x,y) has a [x^2 +(y-5)^2] in it.

When that is equal to zero, T is maximum, T = 36 degrees Celsius.

To make that zero, then x=0 and y=5.

I say the likely location is at (0,5).

iii)Suppose you like to sleep within the temperature range of 20celcious to 25celcious. Where would you put your bed?

I'd solve for area in the room where the temperature is between 20 and 25 degrees Celsius.

When T = 20 degree:

20 = 36 -(1/5)[x^2 +(y-5)^2]

-16 = -(1/5)[x^2 +(y-5)^2]

80 = x^2 +(y-5)^2 ---------------(1)

That is a circle whose center is at (0,5), and whose radius is sqrt(80) = 8.94 ft.

When T = 25 degree:

25 = 36 -(1/5)[x^2 +(y-5)^2]

-11 = -(1/5)[x^2 +(y-5)^2]

55 = x^2 +(y-5)^2 ---------------(1)

That is a circle whose center is at (0,5), and whose radius is sqrt(55) = 7.42 ft.

Therefore, I will place my bed within the area that is 7.42 ft and 8.94 ft away from the heater. The area's inner radius is 7.42 ft; the outer radius is 8.94 ft.

iv)Determine the locations in the room where the temperature is lowest.

They are the points in the room that are farthest from (0,5).

Just by looking, they are (10,10) and (10,0).

The temperature there is 11 degrees Celsius.

They are the points where [x^2 +(y-5)^2] is maximum, where T(x,y) is minimum.

This is where partial derivatives will come in.

Your lecturer's approach.