Question

Business/multivariable calc question help needed asap!!!!

Answer 1

Answer:

There is a   min    value of   800   located at (x,y) =   (16, 12)  

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Explanation:

Let's solve the second equation for y

4x+3y = 100

3y = 100-4x

y = (100-4x)/3

We'll plug that into the first equation

f(x,y) = 2x^2+2y^2

g(x) = 2x^2+2((100-4x)/3)^2

g(x) = 2x^2+(2/9)*(100-4x)^2

g(x) = 2x^2+(2/9)*(10,000-800x+16x^2)

This graphs a parabola that opens upward, due to the positive leading coefficient. This g(x) curve has its vertex point at the minimum.

Apply the derivative to help find the minimum

g(x) = 2x^2+(2/9)*(10,000-800x+16x^2)

g ' (x) = 4x+(2/9)*(-800+32x)

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Set the derivative function equal to 0 and solve for x

g ' (x) = 0

4x+(2/9)*(-800+32x) = 0

4x+(2/9)*(-800)+(2/9)*(32x) = 0

4x-1600/9+(64/9)x = 0

9(4x-1600/9+(64/9)x) = 9*0

36x-1600+64x = 0

100x-1600 = 0

100x = 1600

x = 1600/100

x = 16

Use this x value to find y

y = (100-4x)/3

y = (100-4*16)/3

y = (100-64)/3

y = 36/3

y = 12

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Therefore, (x,y) = (16,12) leads to the largest value of f(x,y) = 2x^2+2y^2

That smallest f(x,y) value is...

f(x,y) = 2x^2+2y^2

f(16,12) = 2*16^2+2*12^2

f(16,12) = 800