Question

if the sum of the two numbers is 16. the sum of their squares is a minimum. Determine the numbers.

Answer 1

You have that if you have two numbers \(x\) and \(y\), you want \(x+y=16\) but you want \(x^2+y^2\) to be a minimum. So, you can write \(y\) as \(y=16-x\).. so you want to minimize:

\[ x^2+(16-x)^2\]
...
Expand this:
\(x^2+(256-32x+x^2)\\
=2x^2-32x+256\)

Now, you can view this as a quadratic function, and so since it has the form \(ax^2+bx+c\), you have that \(a=2>0\), so the shape is like this \(\cup\), and so the vertex is the minimum value. So finding the vertex (or the minimum) will tell you the x-value for which this quadratic function is at a minimum. So, an easy way to do this is to convert your equation into vertex form: \(a(x-h)^k\), where the vertex is \((h, k)\) using the complete the square method.

\(2x^2-32x+256\\
=2(x^2-16x+128)\)

add and subtract \(\left(\dfrac{b}{2}\right)^2=\left(\dfrac{16}{2}\right)^2=64\)
\(=2\left( x^2-16x+64-64+128\right)\\
=2(x-8)^2+64\)

And so the vertex is \((8, 64)\), so the x-value of the minimum occurs at \(x=8\)

And so since \(y=16-x\) from our equation above \(y=16-8=8\)

so \(x=8, y=8\)
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If you know calculus, you can essentially just take the derivative of \(x^2+(16-x)^2\) and set it to 0 , and solve for x

Answer 2

Hm sorry I forgot to carry out the 2 .. you should have \(=2(x-8)^2+128\), giving a vertex of (8, 128) , but the rest of the answer is still the same

Answer 3

I wish I could edit replies... I meant to write the vertex form is \(a(x-h)^2+k\)