Find two numbers whose difference is 8 and whose product is a minimum.

Find two numbers whose difference is 8 and whose product is a minimum.

Let *P* be the product

Let *x* be the first number

Let *y* be the second number.

*x*-

*y*= 8 ---- (1)

*P*=

*xy*----- (2)

Solve for *x* from equation (1): *x* = *y* + 8 -----> I understand this

Sub *y* + 8 for *x* into equation (2): *P* = (*y*+8)*y* = *y*2 + 8*y -> I understand*

Rewrite *P* = *y*2 + 8*y* in the standard form by completing the square. -->I understand

Vertex = (-4, -16). ---> I don't know how they found the vertex?

The product is minimum (*P* = −16) when *y* = −4. ---> I don't understand

Therefore *x* = *y* + 8 = − 4 + 8 = 4. ---> I understand

The two numbers are -4 and 4.

I am pretty confused over this example. Can you thoroughly explain this example to me?

Thanks,

MB

### 2 Answers | Add Yours

Let x and y be the two numbers

We want x-y = 8 and xy=P as small as possible

As you wrote, x=8+y

Once you have that, you can substitute into the other equation, and get

P=(y+8)y = y^2 + 8y

Just as you had.

NOW:

the nice thing about this, is you have P written as a function of only 1 variable, and P is a parabola. Parabolas come in two flavors: rightsideup and upsidedown:

"Right side up" parabolas have a minimum. There is a spot on the parabola that is lower than any other. (The "vertex") But they don't have a maximum. That is, you can always go higher and higher.

"Up side down" parabolas have a maximum. There is a spot on the parabola that is higher than any other (the vertext). But they don't have a minimum. You can always go lower and lower.

Our parabola looks like:

I wish you were my teacher.

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