There is a <u>maximum</u> value of <u>7/6</u> located at (x, y) = <u>(5/6, 7)</u>.
The function given to us is f(x, y) = xy.
The constraint given to us is 6x + y = 10.
Rearranging the constraint, we get:
6x + y = 10,
or, y = 10 - 6x.
Substituting this in the function, we get:
f(x, y) = xy,
or, f(x) = x(10 - 6x) = 10x - 6x².
To find the extremum, we differentiate this, with respect to x, and equate that to 0.
f'(x) = 10 - 12x ... (i)
Equating to 0, we get:
10 - 12x = 0,
or, 12x = 10,
or, x = 5/6.
Differentiating (i), with respect to x again, we get:
f''(x) = -12, which is less than 0, showing f(x) is maximum at x = 5/6.
The value of y, when x = 5/6 is,
y = 12 - 6x,
or, y = 12 - 6*(5/6) = 7.
The value of f(x, y) when (x, y) = (5/6, 7) is,
f(x, y) = xy,
or, f(x, y) = (5/6)*7 = 7/6.
Thus, there is a <u>maximum</u> value of <u>7/6</u> located at (x, y) = <u>(5/6, 7)</u>.
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is 2.
is still 2.