# Calculus - Challenge question

posted by
**Monica** on
.

Our teacher gave us this problem as a challenge.

Some of us have been working on it for a few days

help!

Prove that the largest area of any quadrilateral is obtained when opposite angles are supplementary.

Wow, what a classic and nice question, haven't seen it for years.

My compliments to your teacher for challenging you like that.

I am working at it, ....

Let the sides be a,b,c,and d. Let the angle between a and d be θ, let the angle between b and c be α.

Let x be the diagonal joining the other two angles.

By the Cosine Law: x^2 = a^2 + d^2 – 2ad cosθ and x^2 = b^2 + c^2 – 2bc cos α

Then a^2 + d^2 – 2ad cosθ = b^2 + c^2 – 2bc cos α

Differentiate with respect to θ

2ad sinθ = 2bc sinα (dα /dθ) -----> resulting in dα/dθ = (ad sinθ)/(bc sinα) (#1)

Let A be the area of the quadrilateral, then

A= ½ ad sin θ + ½ bc sin α now take dA/dθ

dA/dθ = ½ ad cos θ + ½ bc cosα(dα/dθ)

= 0 for a maximum of A

So ½ bc cosα(dα/dθ) = -½ ad cos θ

dα/dθ = -(ad cos θ)/(bc cosα) but from (#1) dα/dθ = (ad sinθ)/(bc sin α)

therefore: (ad sinθ)/(bc sin α) = - (ad cos θ)/(bc cosα)

or (sinθ)/(sin α) = - (cos θ)/(cosα)

sinθ cosα = - cos θ sin α

sinθ cosα + cos θ sin α = 0

sin(θ + α) = 0

so: θ + α = 0 , which is not possible, no quardateral or θ + α = 180º or θ + α = 360º also not possible

Therefore the opposite angles must be supplementary, i.e. they must add up to 180º

I can follow it up to the point where you said,

A= ½ ad sin ? + ½ bc sin ?

where does that come form?

Also yur secondlast line is confusing

For any triangle where you know two of the sides a and b and the angle between those 2 sides is A, then the area is ½ab sinA

for you second concern look at sin(θ + α) = 0

Where is the sine of an angle = 0?

at 0,180, 360....

but the angle was θ + α

so

1. θ + α could be 0, but that does not give us a quadrilateral

2. θ + α = 180º

3. θ + α = 360º this is also not possible, since the total of all 4 angles is 360

that leaves us with the inescapable conclusion that θ + α = 180º