anonymous
  • anonymous
one number is 10 more than another number.Form a quadratic function that will allow you to find the minimum product of the two numbers. find the numbers and the minimum product.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@jim_thompson5910
anonymous
  • anonymous
so far x=10+y how to make it into quadratic equation
anonymous
  • anonymous
@mathmale

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anonymous
  • anonymous
@calculusxy any idea
anonymous
  • anonymous
Let \(x\) be the first number. Then the second number is \(x+10\). The product of the two is \(x(x+10)=x^2+10x\). To find the minimum of the product, you can complete the square to write this quadratic in vertex form \((x-a)^2+b\), which will be minimized when \(x=a\) since that makes the first term disappear, leaving you with \(b\).
anonymous
  • anonymous
o wow
anonymous
  • anonymous
|dw:1449275959494:dw|
anonymous
  • anonymous
now solve for x?
anonymous
  • anonymous
Well, strictly speaking you can't solve for \(x\) because this isn't an equation. From your result, you have that the vertex of the parabola is \((a,b)=(-5,-25)\), so \(b=-25\).
anonymous
  • anonymous
so how do i find the minimum im lost
anonymous
  • anonymous
\((x-a)^2+b\) has a minimum value of \(b\) when \(x=a\). Given that you have \((x+5)^2-25\), what value of \(x\) will make the squared term disappear?
anonymous
  • anonymous
-5
anonymous
  • anonymous
Right, so when \(x=-5\), you're left with \(0-25\), which means ...
anonymous
  • anonymous
min value of -25
anonymous
  • anonymous
That's right
anonymous
  • anonymous
Thank you very much:D im trying to teach a friend of mine over the phone how to do this ...
anonymous
  • anonymous
havnt done stuff like this since high school
anonymous
  • anonymous
Thanks again
anonymous
  • anonymous
yw

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