I am preparing for the GRE Math Subject Test. One of the books that I like for this test is GRE Math Subject Test (Advanced) - By Morris Bramson - 1983. It is a very old book, available online but with poor scanning quality. However, I believe, it contains awesome/tricky problems that may be useful to prepare for the GRE.
I am not sure if this question is stated in the right way or not.
Find the minimum value of the function $f(x,y)=x^3+y^3-3xy$.
$\text{(A) 0} \space\space\space\space\space\space\space\space\space\space \text{(B) -1} \space\space\space\space\space\space\space\space\space\space \text{(C) -2} \space\space\space\space\space\space\space\space\space\space \text{(D) 1} \space\space\space\space\space\space\space\space\space\space \text{(E) 2}$
Originally from the book:
I know the way to solve easy max/min problems.
$f_x=3x^2-3y$, and $f_y=3y^2-3x$. Equating these partial derivatives to zero and solving the system of equations, we get
$(x,y)=(1,1)$ as a critical point. Now evaluating $f(1,1)=1^3+1^3-3(1)(1)=-1$. Hence $\text{(B)}$ is correct (as mentioned in the book).
However, this is not "the minimum". This is "a" local minimum. The function has no minimum because we can set $y=0$ and $x$ as small as we like.
Excuse me for my poor English (I am Arabic). What do you think? Am I right that the function has no minimum and that the question is not stated in a right way?
Thanks.
Let's $f = x^3 +y^3-3xy$.
Then, $$f = x^3 +y^3+1^3 -3 \times x \times y \times 1 -1\\ \Rightarrow f = (x+y+1)(x^2+y^2+1-xy-x-y)-1 \\ \Rightarrow f = \frac{1}{2}(x+y+1)((x-y)^2+(x-1)^2+(y-1)^2) -1 $$
To minimize $f$, the term $(x-y)^2+(x-1)^2+(y-1)^2$ must be minimized as it is always $\geq 0$ i.e.
$$(x-y)^2+(x-1)^2+(y-1)^2\geq 0$$ $$ \Rightarrow min((x-y)^2+(x-1)^2+(y-1)^2) = 0$$ As $(x-y)^2 \geq 0$, $(x-1)^2 \geq 0$ and $(y-1)^2 \geq 0$, the above equation splits into $$ \Rightarrow \bigg( min \big((x-y)^2 \big) +min \big((x-1)^2 \big) +min \big((y-1)^2 \big) \bigg) $$ $$ \Rightarrow \bigg( x-y=0, x-1=0 \text{ and } y-1=0 \bigg) $$ $$ \Rightarrow x=y=1 $$ so that $(x-y)^2+(x-1)^2+(y-1)^2 = 0$ which is its minimum value.
This minimum value will be achieved by making $(x-y)^2=0$, $(x-1)^2=0$ and $(y-1)^2 = 0$.
Thus, $min(f) = \frac{1}{2}(1+1+1) \times 0 -1 = -1$