There is a maximum value of 6 located at (x, y) = (1, 6).
The function given to us is f(x, y) = xy.
The constraint given to us is 6x + y = 12.
Rearranging the constraint, we get:
6x + y = 12,
or, y = 12 - 6x.
Substituting this in the function, we get:
f(x, y) = xy,
or, f(x) = x(12 - 6x) = 12x - 6x².
To find the extremum, we differentiate this, with respect to x, and equate that to 0.
f'(x) = 12 - 12x ... (i)
Equating to 0, we get:
12 - 12x = 0,
or, 12x = 12,
or, x = 1.
Differentiating (i), with respect to x again, we get:
f''(x) = -12, which is less than 0, showing f(x) is maximum at x = 1.
The value of y, when x = 1 is,
y = 12 - 6x,
or, y = 12 - 6*1 = 6.
The value of f(x, y) when (x, y) = (1, 6) is,
f(x, y) = xy,
or, f(x, y) = 1*6 = 6.
Thus, there is a maximum value of 6 located at (x, y) = (1, 6).
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The provided question is incomplete. The complete question is:
"Find the extremum of f(x, y) subject to the given constraint, and state whether it is a maximum or a minimum. f(x, y)=xy; 6x+y=12.
