Calculus 2: The function f(x,y) = ax^2 + bxy + cy^2 has an average value of 20 on the square
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Calculus 2: The function f(x,y) = ax^2 + bxy + cy^2 has an average value of 20 on the square

[From: ] [author: ] [Date: 12-04-28] [Hit: ]
b) Find two different choices for f with average value 20 on the square, and sketch their contour diagrams.∫∫f(x,--> you have a square region therefore it doesnt matter what order you choose.........
The function f(x,y) = ax^2 + bxy + cy^2 has an average value of 20 on the square 0 <= x <= 2,0 <= y <= 2

a) what can you say about the constants a, b, c?

b) Find two different choices for f with average value 20 on the square, and sketch their contour diagrams.

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a)

Just take the double integral:

∫∫f(x, y)dxdy
--> you have a square region therefore it doesn't matter what order you choose...let's do the x-integral first

∫f(x, y)dx = ax³/3 + bx²y/2 + cxy² + C
--> evaluate from x = 0 to x = 2

8a/3 + 4by/2 + 2cy² = 8a/3 + 2by + 2cy²
--> now do the y integral

8ay/3 + by² + 2cy³/3 + C
--> evaluate from y = 0 to y = 2

16a/3 + 4b + 16c/3

This is the value of the integral, for the average to be 20, then this amount divided by the area of region: (2*2 = 4):

average = (16a/3 + 4b + 16c/3) / 4 = 4a/3 + b + 4c/3 = 20

So you can say that the values a, b, and c, lie on a plane given by:

4a + 3b + 4c = 60

b) I can't (and don't want to) draw contour maps on YA!

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∬f(x,y)dA / area of square gives you the average of f on the square

=> ∬fdA = 80 => ∫[0 to 2] ∫[0 to 2] ax^2 + bxy + cy^2 dydx = 80

=> ∫[0 to 2] 2ax^2 + 2bx + 8c/3 dx = 80

=> relationship between the three constants 16c/3 + 16a/3 + 4b = 80

b) Let b = 0, a = 0, then c = 15 => f(x,y) = 15y^2, which is a cylinder
Let c = 0, b = 0, then a = 15 => f(x,y) = 15x^2, which is a cylinder
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