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

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maythany | (Level 3) eNoter

Posted December 13, 2012 at 12:03 AM via web

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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) Pxy ----- (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 = y2 + 8y -> I understand
Rewrite P = y2 + 8y 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

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted December 13, 2012 at 12:39 AM (Answer #1)

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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:





We want to find that spot at the bottom (the red thing is a makeshift arrow).  That is where P is at its lowest






`P=ay^2+by+c = 1 y^2 + 8y + 0`

One way to find the vertex is to use the formula:

`(-b)/(2a)`

This gives you the first coordinate of the vertex.  In our case

`(-8)/(2*1) = -4`

To get the second coordinate, plug the first one into the parabola equation.  In our case:

`P = (-4)^2 + 8 (-4) = -16`

So the vertext is (-4,-16)

So, when y = -4, then P = -16 (and x = y+8 = 4)





Completing the square is a little harder, but it is another way to find the vertex.

Completing the square means you want to write the equation in the form:

`P = a(y-h)^2 + k`

The nice thing about writing P this way is that the vertex is (h,k)

That is, if you have the equation written in standard form, you can tell at a glance what the vertex is.

If you have that done, great.  Just in case:



We want:

`y^2 + 8y = a(y-h)^2 + k`

The "a" is the same in both cases.  That is, for us,

`P=ay^2+by+c = 1 y^2 + 8y + 0`

The "a" here is 1, so we have:

`P = 1(y-h)^2 + k`


So, we want:


`y^2 + 8y = (y-h)^2 + k`

Expand the right hand side:

`y^2 + 8y = (y-h)(y-h) + k = y^2 - yh - yh + h^2 + k`


So:

`y^2 + 8y + 0 =  y^2 - (2h)y + h^2 + k`

We want the two sides to be equal.

So we need -2h = 8, and we need  `0=h^2+k`

So we need h = -4

If h=-4, then `0=h^2+k=(-4)^2+k = 16+k`

So we need k=-16


So, if we plug these back into what we started with, we have:

`y^2 + 8y = (y-h)^2 + k`

`y^2 + 8y = (y+4)^2 -16`


This is standard form.  The vertex is (h,k) which is (-4,-16)
 (you take the negative of what is in parentheses)

That means, the minimum possible P is -16, and it happens when y=-4
(which means x must be 4)

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maythany | (Level 3) eNoter

Posted December 17, 2012 at 1:57 AM (Answer #2)

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I wish you were my teacher.

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