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
