Quadriateral Max/Min Question

[BayernBlues]

1. The problem statement, all variables and given/known data

The diagonals of a quadrilateral are perpendicular. The sum of the lengths of the diagonals is 6cm. What is the maximum area of such a quadrilateral?

2. Relevant equations



3. The attempt at a solution

let x and y be the lengthes of the diagonals. Then the area of the quadrilateral is calculated by:

a = ½ * x * y

From the given conditions you know: x + y = 6 ==> y = 6 - x

Plug in the term for y into the first equation:

a(x) = ½*x*(6 - x) = -½x² + 3x


I do not know what to do from here. I know that there is a graphing method but I'd rather do it through differentiation so could someone do the solution that way. The answer should be y=3 and x=3 so the dimensions are 3 x 3 I think.

[neutrino]

Do you know how to find the extrema(maxima and minima) of a given function with respect to a variable?

[BayernBlues]

I'm more confused now. If I differentiate a(x), this is what happens
a'(x)=2/2 x + 3
0= x + 3
x= -3

y= 6 - x
y = 6 - (-3)
y= 9

So the dimensions are 9 x 3? That doesn't seem right.

[neutrino]

You missed the minus sign.

[BayernBlues]

6- -3 is 9.

I'm going over it and that's still the only answer I get for some reason.

[neutrino]

I meant when you differentiated the original equation.