The basic technique for dealing with each of these problems is that of completing the square.
The first one, \(\displaystyle t^{2}-9t-36\).
Take a half of the coefficient of the linear term, (the 9t term), -9/2, square it, 36/4, add it to the expression and subtract it as well.
That gets you
\(\displaystyle t^{2}-9t +\frac{36}{4}-36-\frac{36}{4}\), which is still equal to the original,
and which can be written as
\(\displaystyle \left(t-\frac{9}{2}\right)^{2}-45\) .
The minimum value for that will be -45 occurring when t = 9/2, since that makes the expression inside the bracket zero. For any other value of t, the squared term will be greater than zero, making the whole expression greater than -45.
For the method to work, the coefficient of the squared term has to be 1. If the coefficient is not equal to 1, it has to be removed as a factor.
For example, if we had, (not one of yours),
\(\displaystyle -2x^{2}-8x+17\),
we would proceed as follows.
\(\displaystyle -2(x^{2}+4x)+17=-2(x^{2}+4x+4-4)+17\)
\(\displaystyle =-2\{(x+2)^{2}-4\}+17\) ,
\(\displaystyle =-2(x+2)^{2}+8+17\\=25-2(x+2)^{2}\)
Looking at that, its maximum value will be 25, occurring when x = -2, (since that makes the expression inside the bracket equal to zero).
The two variable example further down, \(\displaystyle x^{2}+y^{2}-4x+2y\), can be dealt with in much the same way.
Write it as
\(\displaystyle x^{2}-4x+y^{2}+2y\) ,
and complete the square on x and y separately. It's then easy to see what the minimum value is, and the values of x and y that give that value.
Tiggsy.
