Determine two numbers whose difference is -4 and whose product is a minimum. Let x be one number and let y be the other.
There is another question similar to this but with maximum instead of minimum, the concept should be the same, right?
Determine two numbers whose difference is -4 and whose product is a minimum. Let x be one number and let y be the other.
There is another question similar to this but with maximum instead of minimum, the concept should be the same, right?
I am a newbie myself, but I think that one way to solve this is to think something along these lines.
If y is negative, the product y(y-4) will be positive.
If y is positive, there will be a few numbers for which the product y(y-4) is negative, and some for which the product will be positive.
In this particular case it is easy to manually figure out for which values of y the product will be negative, since there are only so many positive values of y for which (y-4) will remain negative.
If we consider $\displaystyle ax^2+bx+c$ completing the square gives $\displaystyle a\left(x-\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a}$ which when $\displaystyle a>0$ we may clearly see that $\displaystyle a\left(x-\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a}\geqslant c-\frac{b^2}{4a}$ so that it attains it's minimum at $\displaystyle x=\frac{b}{2a}$. Geometrically all we did was find the vertex of the parabola.
Actually, no. There will be an infinite wet of numbers for which the product y(y- 4) is negative. You cannot assume the number will be a whole number. Completing the square, as Drexel28 suggested, will always give you the minimum or maximum point (vertex of the parabola) of a quadratic function.
Again, no, there are an infinite number of them.In this particular case it is easy to manually figure out for which values of y the product will be negative, since there are only so many positive values of y for which (y-4) will remain negative.