Expert: Paul Klarreich - 3/28/2006

Question
i'm denise and i'm doin calculus.i made an equation in order to use the differentiation method to solve a problem but it din't work.if the hypotenuse of a right angle trianlge is ten,what must the sides be so that the area is a maximum ?

Answer
Hi, Denise,

First suggestion:  When you enter your problems into the computer, they are easier for me to read if you:

1. Use those big buttons on the left and right sides of the keyboard -- the ones that change  'this' into 'This', and 'i' into 'I'.

2. Add a couple of spaces after a period.  That way, I can see where a new sentence begins.

3. If you did something, like 'made an equation', send the equation along.  Then I can see what you are doing wrong and possibly correct it.  Every little bit helps.

For your maximization problem, your general scheme should be:

0. (If appropriate) Draw a clear and accurate diagram.  No raw scribbles, please -- a good diagram.
1. Select some variable.  
2. Express the quantity to be maximized in terms of that variable.
3. Differentiate, set it = 0, solve, etc.

Of course, #3 may be a lot of work, but 0,1,2, are the hard parts because they involve thinking and using your life experience.  (a.k.a. general mathematical knowledge)

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

In this case, how about this:

     /|
    / |
10  /  |
  /   | y
 /    |
/     |
- - - -+
   x
suppose we let one leg of the right triangle be x and the other leg be y.  We want to maximize A = xy.  (By the way, do you live in a country where I must write 'maximise'?)

But A = xy expresses A in terms of two variables.  That's N.G.  So we need a way to relate x and y to each other.  But of course, x and y are legs of a right triangle.  So:

x^2 + y^2 = 100

and  y = sqrt(100 - x^2)

You will put that into the A = xy and get something giving A in terms of x alone, and you are on your way.

HOWEVER, this old geezer is basically lazy and hates things with radicals in them.  So he uses this trick:

To maximize (ise?) A, we could maximize A^2 instead.  Same thing.

A^2 = x^2 y^2 = x^2(100 - x^2) = 100x^2 - x^4

[I multiply that out because I hate the product rule, too.]

Now we are on our way:

dA^2
---- = 200x - 4x^3
dx

Set that equal to zero.

200x - 4x^3 = 0
4x(50 - x^2) = 0

x = 0,  x = +- sqrt(50) = +- 5 sqrt(2)

Naturally, the zero is a minimum, not a maximum, and the negative solution does not apply.

Also,  y = sqrt(100 - x^2) = sqrt(100 - 50) = sqrt(50) = 5 sqrt(2), which is the same as x.  In other words, this will be an isosceles triangle.  You guessed that, didn't you?