Question

find extreme values of (x^3y^2)(12-3x-4y)​

Answer 1

The extrema = (2, 1).

• Any position in calculus where the value of a function is either greatest (a maximum) or smallest (a minimum) is known as an extremum, plural of Extrema. There are maxima and minima that are both absolute and relative (or local).
• At an absolute maximum, the value of the function is greater than any other point in the period of interest, whereas at a relative maximum, it is greater than its value at points that are immediately adjacent. If the function is smooth rather than peaked at relative maxima inside the interval, its rate of change, or derivative, is zero. At a position where the function has neither a maximum nor a minimum, the derivative could be zero.

Given that, f(x) = $(x^{3} y^{2} )(12-3x-4y)$

We first need to find the partial derivatives, after which the critical points can be found out.

This function attains the extrema at (2,1).

Hence, the extrema = (2, 1).

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