Question

find two numbers whose difference is -30 abd whose product is a minimum

Answer 1

Let x and y be the numbers.
x-y=-30
Product = xy = (-30+y)-y

Answer 2

x-y=-30 so x=-30+y
Product = xy = (-30+y)*y which is a quadratic.

Quadratic equations always have their minimum or maximum at the vertex (notice that this one is an upward opening parabola, so has a minimum).

Find the vertex, and you have your y that minimizes the product; substitute that back into x=-30+y

Answer 3

@DebbieG can you continue it, i can understand it little, sorry if i ask you so

Answer 4

Well, I think I would have opted to solve the difference equation for y, instead of for x.... just so that you end up with a quadratic equation with y as a function of x, which is how we are typically used to seeing it.

so let's start over. If the two numbers are x and y, you have:

x-y=-30 so y=x + 30 (do you understand that step?)

Now what we want to minimize is the product, xy. So substituting in the expression we obtained for y above, we have:
Product = \(\large xy = x(x + 30) = x^2 + 30x\)

So now the job is to find the x-value that minimizes \(\large f(x) = x^2 + 30x\)

Since that is a quadratic function, it's graph is an upward opening parabola. So find the vertex, and you have found the minimum value of the function. Thus the x-coordinate of the vertex is where the function is at a minimum.

Use the formula for the x-coordinate: \(\large x=\dfrac{-b}{2a}\) to find that x-coordinate.

Once you have x, that is one of the numbers. Use the fact that \(\large y=x + 30\) to find y. And there are the two numbers.