Question

6. Let px -5 and Py-2. Consi 2. Consider the following equation (a) Find the partial derivative with respect to x, and call it fr (b) Find the partial derivative with respect to y, J^ and call it fu (c) Plug in the above to get equation 1 (and simplify) Equation 1L. (d) Find 4 (x, y) points such that f(x, y) - 80 (e) Any combination of (x, y) that leads to f(x,y) 80 must satisfy what equation? Let this equation be equation #2 (f) Find the point at the intersection of equation #1 and equation #2

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Answer #1

f(x,y) = 2xy

a)

f_{x} = rac{partial 2xy}{partial x} = 2y

b)

f_{y} = rac{partial 2xy}{partial y} = 2x

c)

rac{f_x}{f_y} = rac{p_x}{p_y}

rac{f_x}{f_y} = rac{2y}{2x} = rac{y}{x}

rac{p_x}{p_y} = rac{5}{2}

rac{y}{x} = rac{5}{2} => 5x = 2y --------> equation 1

d)

f(x,y) = 80 => 2xy = 80 => xy =40 => y=rac{40}{x}

We can assume different values for x. Substituting in the above equation, we can get the corresponding values of y.

i. x =1 => y =40

ii. x =2 => y =20

iii. x =4 => y =10

iv. x =5 => y =8

Therefore, 4 sets of (x,y) satisfying f(x,y)= 80 can be (1,40), (2,20), (4,10), (5,8)

e)

Any combination of (x,y) which satisfies f(x,y) =80 must satisfy,

2xy =80

i.e, xy =40

xy =40 -------------> Equation 2

f)

5x = 2y --------> equation 1

xy =40 -------------> Equation 2

From equation 2,

y = rac{40}{x}

Substituting this in equation 1,

5x = 2* rac{40}{x} = rac{80}{x}

=> 512-80

=> x^2 = 16

=> x = pm 4

=> y = pm 10

That is, equation 1 and equation2 intersects at 2 points, (x,y) = (4,10) and (-4,-10)

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