Find the minimun value of the product p of two numbers, one of which is 9 less than twice the other.
Denote the first number by x. Then the second will be
2*x - 9
("9 less than twice the other").
Its product p is `x*(2x-9) = 2x^2 - 9x.`
It is not difficult to find the minimum value of this expression (and to show that it has a minimum). For this we select a perfect square:
`2x^2 - 9x = 2*(x^2 - (9/2)*x) = 2*(x - 9/4)^2 - 2*(9/4)^2.`
Here I use the formula `(a-b)^2 = a^2 - 2ab + b^2.`
The first summand, `2*(x - 9/4)^2,` cannot be less than zero. Moreover, it is zero if and only if x = 9/4. So the minimum value of the initial expression is
`-2*(9/4)^2` = -81/8 = -10.125. This is the answer.
P.S. The numbers which form this product are x=9/4 and 2x-9=-9/2.