What is the greatest possible area of a triangular region

[RonPurewal]

thanks christiancryan. I too, answered the question wrong (marked obvious A). And the way you explained the question, I now know a very important concept of working the maxima-minima without the differential, and the beauty is that the concept can be used everywhere from number system to coordinate geometry.

good.

also, don't forget that the only mathematics required for this exam is the stuff up to and including high-school algebra and geometry. therefore, if you find yourself employing calculus, trigonometry, etc. in an attempt to solve one of these problems, you can rest assured that you are overcomplicating the issue.

[sergiolopez]

I tried in this way:

First I calculated the maximum square that could fit in a triangle. I know that the diagonal=2 then the side S=2/SQRT(2) and as a consequence the area is S^2=2. The triangle is 1/4 of the area of the square --> A=1/2

Answer is B

[Abhilash Paul]

I did it this way. Area of a triangle can also be = 1/2 * ab * cos(x)
where a, b are two sides of the triangle, and x is the included angle.
Max area is when cox(x) is max (cos(x) is max for x= 90 deg and = 1)
So the answer is 1/2 *ab = 1/2 if you take the two sides as the radius of the circle, =1 in this case.
This explanation may prob be out of the scope of GMAT, not sure but this sure was a very fast method.

[RonPurewal]

I did it this way. Area of a triangle can also be = 1/2 * ab * cos(x)
where a, b are two sides of the triangle, and x is the included angle.
Max area is when cox(x) is max (cos(x) is max for x= 90 deg and = 1)
So the answer is 1/2 *ab = 1/2 if you take the two sides as the radius of the circle, =1 in this case.
This explanation may prob be out of the scope of GMAT, not sure but this sure was a very fast method.

no trigonometry is required on the gmat, but, on one out of several hundred problems, it may help.

[psps]

Here is an easy solution, but the prose is a bit messy to write down. Call the triangle ATC with AC being the base. Let A and C be the two foci of an ellipse placed on the X-axis in a rectangular coordinate system.

Clearly, any triangle ATC can be circumscribed by an ellipse, where T is the movable point which traces the curve. Since by the definition of an ellipse AP + CP is constant, and by the fact that the area of a triangle equals one-half the base times the height, it is obvious that the maximum area of a triangle with a given base must be isosceles.

With this is mind we can now designate the base (AC) as "˜b’ and the two sides (AT, CT) as "˜s’.

For the perimeter (P) we then have the P=2*s +b. And for the height (H) we have (using Pythagoras) H= sqrt [s^2-(b/2) ^2]

From P= 2*s+b, we have
s= (P-b)/2, and H = sqrt [([P-b]/2) ^2 + (b/2) ^2]

So the area can now be written as
Area = 1/2 * b * sqrt [([P-b]/2) ^2 + (b/2) ^2]

Since P is constant, to find the maximum area, we simply differentiate dA/db and set it to zero. From here we get that P= b/3 and if follows that the triangle is equilateral.

Pinchas

[jlucero]

I wouldn't exactly call that an easy solution :) For those of you lucky enough to have a good recollection of trigonometry, you might be able to use it on the GMAT, but as several other instructors have mentioned- you shouldn't try to learn it to take the GMAT.

The problem with your elipse analogy is that it doesn't fit with this problem. If A is the center of our circle and C is on the circle, point T can be anywhere around point A (a circle) rather than around both A and C (an ellipse).

As other instructors have mentioned, if the base of this triangle must be 1, then the formula 1/2bh means we simply need to maximize the height to maximize the area.