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Given that $x + y = -1$ find the largest value of $xy$
(a) by co-ordinate geometry
(b) by calculus
(c) by algebra.
Here are some solutions from Koopa,Boston College, USA. Vassil, Lawnswood Sixth Form, Leeds sent in similar methods. Can you find a co-ordinate geometry (i.e. graphical) method or yet another different method?
Method 1
We have $x + y = -1$. So, to maximise $xy$, I need to maximize
\[-x(x + 1) = -[(x + 1/2)^2 - 1/4] = -(x + 1/2)^2 + 1/4,\]
so, $xy$ is maximized at $x = -1/2$ and the maximum value is $1/4$.
Method 2
Let $f(x) = -x(x + 1)$, then by differentiation $f'(x) = -2x - 1$ and to find a maximum or minimum $f'(x)= 0$ gives $x = -1/2$. The second derivative test easily verifies that this indeed gives a maximum so the maximum value is $1/4$.
Method 3
By the AM-GM inequality, we have
\[(xy)^{(1/2)} \leq (x + y)/2\]
so \[xy \leq (-1/2)^2 = 1/4.\]
Three Ways
Age 16 to 18
Challenge Level 
Given that $x + y = -1$ find the largest value of $xy$
(a) by co-ordinate geometry
(b) by calculus
(c) by algebra.
Here are some solutions from Koopa,Boston College, USA. Vassil, Lawnswood Sixth Form, Leeds sent in similar methods. Can you find a co-ordinate geometry (i.e. graphical) method or yet another different method?
Method 1
We have $x + y = -1$. So, to maximise $xy$, I need to maximize
\[-x(x + 1) = -[(x + 1/2)^2 - 1/4] = -(x + 1/2)^2 + 1/4,\]
so, $xy$ is maximized at $x = -1/2$ and the maximum value is $1/4$.
Method 2
Let $f(x) = -x(x + 1)$, then by differentiation $f'(x) = -2x - 1$ and to find a maximum or minimum $f'(x)= 0$ gives $x = -1/2$. The second derivative test easily verifies that this indeed gives a maximum so the maximum value is $1/4$.
Method 3
By the AM-GM inequality, we have
\[(xy)^{(1/2)} \leq (x + y)/2\]
so \[xy \leq (-1/2)^2 = 1/4.\]
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