Answer:
There is a min value of 376 located at (x,y) = (9,7)
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Explanation:
Solve the second equation for y
x+y = 16
y = 16-x
Then plug it into the first equation
f(x,y) = 3x^2+4y^2 - xy
g(x) = 3x^2+4(16-x)^2 - x(16-x)
g(x) = 3x^2+4(256 - 32x + x^2) - 16x + x^2
g(x) = 3x^2+1024 - 128x + 4x^2 - 16x + x^2
g(x) = 8x^2-144x+1024
The positive leading coefficient 8 tells us we have a parabola that opens upward, and produces a minimum value (aka lowest point) at the vertex.
Let's compute the derivative and set it equal to zero to solve for x.
g(x) = 8x^2-144x+1024
g ' (x) = 16x-144
16x-144 = 0
16x = 144
x = 144/16
x = 9
The min value occurs when x = 9. Let's find its paired y value.
y = 16-x
y = 16-9
y = 7
The min value occurs at (x,y) = (9,7)
Lastly, let's find the actual min value of f(x,y).
f(x,y) = 3x^2+4y^2 - xy
f(9,7) = 3(9)^2+4(7)^2 - 9*7
f(9,7) = 376
The smallest f(x,y) value is 376.
