Find the dimensions of the isosceles triangle of maximum area if the perimeter is to be 24 cm.?

Answer 1

The answer is an equilateral triangle, of course, but I'd imagine getting the answer by a maximizing equation is the point to this problem.

Let b be the base, s be the congruent sides. Then:

b + 2s = 24
2s = 24 - b
s = 12 - b/2

The area of a triangle is bh/2, where h is the altitude. Since the altitude of an isosceles triangle bisects the base, it follows from the Pythagorean Theorem that h = sqrt(s² - (b/2)²). So, we now have the formula that needs to be maximized:

bh/2 = (b*sqrt(s² - (b/2)²)/2

Substitute s = 12 - b/2 in the above formula to eliminate s:

(b * sqrt((12 - b/2)² - (b/2)²)/2
(b * sqrt(144 - 12b + b²/4 - b²/4))2
(b * sqrt(144 - 12b))/2

So that's the formula, simplified as much as possible. To maximize this, take the derivative. (This is an exercise in itself - it uses the multiplication rule and the chain rule.) The derivative I calculate is:

(sqrt(144 - 12b) + b(-12)/2sqrt(144 - 12b))/2
(sqrt(144 - 12b) - 6b/sqrt(144 - 12b))/2

Getting a common denominator, this turns into

(144 - 12b - 6b)/sqrt(144 - 12b))/2
(72 - 6b - 3b)/sqrt(144 - 12b)
(72 - 9b)/sqrt(144 - 12b)

Finally, we see that, for the derivative to be zero, b = 8 is required. Phew - a lot of work to reach a conclusion we knew at the beginning!

Answer 2

equilateral.
all sides are 8
height is 4 root 3