Completing the square problem

Hi all,

I'm trying to brush up on my maths by working through an A-level textbook; so far I've learned a lot (it's been a long time since I did the subject for GCSE). I'm not finding it easy as maths has never been my strong subject, but with perseverence I hope I'll half-understand it!

There's one problem set in the book which seems to me to have been introduced a bit out of the blue and so far it's left me a bit stumped... I was wondering if anyone could offer a pointer on this. The problem is:

A piece of wire has length 20cm and it is bent into a rectangle with one side equal to x cm. Show that the area of the rectangle, A cm^2, is given by A = x(10 -x)...

(the next part of the problem is to complete the square on the expression for A and hence find the greatest value of A and the value of x for which this occurs).

The only idea I had was to multiply it out to:

A = 10x - x^2

but I can't see how this can bring me any closer to the solution (which I did look at in the back of the book)... Any help would be greatly appreciated!

Reply 1

TenOfThem

-(x-5)^2 = -x^2 + 10x - 25

A = -(x-5)^2 + 25 = 25 - (x-5)^2

Since

(x-5)^2

is 0 or bigger

The biggest value for A is 25

(edited 11 years ago)

Reply 2

Hasufel

A=25-(x-5)^2

Reply 3

jonnburton

Thanks a lot for your really quick replies!

Just a couple of things, if you'll forgive my stupidity...

The first is, where do you get -(x-5)^2 from?

And the other is that I would have expanded that out as: x^2 +10x + 25

Where would I be going wrong with that?

Basically I can't see the step from A = x(10 -x) to anything involving x-5

Cheers,

Reply 4

alicebc

You know that the perimeter is 20, so length of one long side + length of one short side = 10. This is because there are 2 sets of equal sides (2x+2y=20, so x+y=10).

As the area is the long side multiplied by the short side it must be x multiplied by y, and if you rearrange x+y=10 you will get y=10-x so another way of writing the area is x(10-x).

You were right to multiply it out, but completing the square is where you rearrange the equation into the form (x-a)^2 + b. So for your equation this will be (x-5)^2 - 25 <-- easiest way to think about this will be to try and produce the 10x by the value of a, and then balance the rest by the value of b. The expansion of (x-5) will give you x^2 -10x+25, so you get rid of the 25 by making the negative value of this the value of b.

When you have an equation in the form y= (x-a)+b you know that the max or min value (I think it can be the minimum sometimes but not sure) will be at (a,b). So in your case at (5,-25). As you know x is the side length and y=A as A is also =(x-5)^2 -25, then the max value for A will be 25 (can't have -ve area), where x = 5.

Another way to do this would be using differentiation. Hope this has helped, and that I haven't gone wrong anywhere!

(edited 11 years ago)

Reply 5

TenOfThem

Original post by jonnburton

Thanks a lot for your really quick replies!

Just a couple of things, if you'll forgive my stupidity...

The first is, where do you get -(x-5)^2 from?

Because I know that

(x-a)^2 = x^2 - 2ax + a^2

And I can see that I need

x^2

and 10x

And the other is that I would have expanded that out as: x^2 +10x + 25

me being dim ... corrected

Reply 6

jonnburton

OK, I think I'm getting this now! Thank you very much for everyone's replies, they've been really helpful!

Reply 7

alicebc

Just realised why I ended up with a negative area haha, I multiplied everything by -1 when I was completing the square and I shouldnt have because the area was not = 0. So if I just multiply area by -1 after I complete the square I get 25. Sorrrrry!