1. ## Quad Functions Max/Min Problem

A number is 7 less than double a second number. The sum of the squares of these two numbers is a minimum. Determine the two numbers.

Need help assigning variables and basically writing the quadratic function. I should be able to complete the square myself though and answer the question.

Any help is appreciated.

2. Originally Posted by Raj
A number is 7 less than double a second number. The sum of the squares of these two numbers is a minimum. Determine the two numbers.

Need help assigning variables and basically writing the quadratic function. I should be able to complete the square myself though and answer the question.

Any help is appreciated.
Let the first number be $x$ and the second be $y$.

Then we are told that $x=2y-7$, then we are told that $x^2+y^2$ is a minimum, so $(2y-7)^2+y^2$ is a minimum.

RonL

3. Alright i dont know what im doing wrong

$x = y^2 + (2y-7)^2$

$x = 5y^2 - 28y + 49$

Then i got
$x = 5(y-14/5)^2 + 49/5$

4. Originally Posted by Raj
Alright i dont know what im doing wrong

$x = y^2 + (2y-7)^2$

$x = 5y^2 - 28y + 49$

Then i got
$x = 5(y-14/5)^2 + 49/5$

x = y^2 +(2y -7)^2 ---------------(1)
No.
Remember that x = 2y -7
And, the sum of x^2 +y^2 is not x.

The righthand side (RHS) of (1) will be a parabola in y.
Normally, a parabola in y is a "horizontal" parabola in that the axis of symmetry is horizontal.
Then, if you are used to minimum as the lowest vertically, you might get lost with a horizontal parabola because its "lowest" is the point that is the leftmost.
So we change the 2nd number from y to x. The first number is y.
And the sum is s.

Thus,
y = 2x -7

s = y^2 +x^2
s = (2x-7)^2 +x^2
s = 5x^2 -28x +49 ----------------(i).
Now the parabola is "vertical", that opens upward. So its vertex is the lowest point, which represent a minimum.

When you do the "completing the square" on (i), you are trying to transform (i) into the form
s = a(x-b)^2 +c ------(ii)
which is still a vertical parabola.
But this time, (ii) shows already the vertex.
The vertex is (b,c).
Or, b is the value of x at the vertex.

c is the value of s at the vertex.

We are looking for x and y when s is minimum---at the vertex.

s = 5x^2 -28x +49
s = 5[x^2 -(28/5)x +(14/5)^2] -5(14/5)^2 +49
s = 5[(x -(14/5))^2 ] -196/5 +49
s = 5[(x-(14/5)^2] +(49/5)

Hence, x = 14/5 at the vertex.
And minimum s = 49/5.

Hence, at the vertex also, y = 2x -7 = 28/5 -7 = (28 -35)/5 = -7/5

Check,
y^2 +x^2 = miminum
(-7/5)^2 +(14/5)^2 =? 49/5
49/25 +196/25 =? 49/5
245/25 =? 49/5
(49*5)/(5*5) =? 49/5
49/5 =? 49/5
Yes, so, OK.

Therefore, the two numbers are -7/5 and 14/5. ---------------answer.

5. tyvm ticol for clarifying the variables