Science:Math Exam Resources/Courses/MATH105/April 2010/Question 01 (c)

MATH105 April 2010

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Question 01 (c)
Find all point(s) (x,y) where $\displaystyle f(x,y)=x^{3}y-8y-5x$ may have a relative maximum or minimum.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How do we define critical points for functions of two variables?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For $f(x,y)$ to have a local maximum or minimum at a point $(a,b)$, that point needs to be a critical point, that is $\displaystyle f_{x}(a,b)=f_{y}(a,b)=0$ For the function $\displaystyle f(x,y)=x^{3}y-8y-5x$ we have the partial derivatives $\displaystyle f_{x}=3x^{2}y-5\quad {\text{and}}\quad f_{y}=x^{3}-8$ To look for critical points, we need to find points which yield a zero in both partial derivatives. Looking in particular at the partial derivative with respect to y we observe that if $\displaystyle f_{y}=0$ then $\displaystyle x^{3}-8=0$ and so $\displaystyle x=2.$ Taking this value of $x$ and substituting it into the equation for the partial derivative with respect to $x$ yields $\displaystyle 3\cdot 2^{2}y-5=0$ so $\displaystyle 12y=5$ or $\displaystyle y=5/12$ We find only one critical point, $(a,b)=\left(2,{\frac {5}{12}}\right).$ Since in order to be a relative maximum or minimum, the point must be a critical point, we have found that the only point that may be a maximum or minimum is (2,5/12). Continuing the Problem: If we assume the second derivative test is applicable (sometimes it is inconclusive), we need $\displaystyle f_{xx}(a,b)f_{yy}(a,b)-f_{xy}^{2}(a,b)>0$ to have a relative maximum or minimum. If we compute the second derivatives, we find $\displaystyle {\begin{aligned}f_{xx}(x,y)&={\frac {\partial }{\partial x}}(3x^{2}y-5)=6xy\\f_{yy}(x,y)&={\frac {\partial }{\partial y}}(x^{3}-8)=0\\f_{xy}(x,y)&={\frac {\partial }{\partial y}}(3x^{2}y-5)=3x^{2}\end{aligned}}$ and so $\displaystyle {\begin{aligned}f_{xx}(a,b)&=6\cdot 2\cdot {\frac {5}{12}}=5\\f_{yy}(a,b)&=0\\f_{xy}(a,b)&=3\cdot 2^{2}=12\end{aligned}}$ Thus, $\displaystyle f_{xx}(a,b)f_{yy}(a,b)-f_{xy}^{2}(a,b)=0-12^{2}=-144$ Since this is a negative number, this critical point is a saddle point and so the conclusion is that the critical point is not a relative maximum or minimum and since there is only one critical point, there are no relative maxima/minima for $f(x,y)$.

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