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Problem:  You are given an isosceles triangle of area 1 with equal sides of length x, height h, and base b.

Part 1: Express the perimeter, P, in terms of x and b.

Answer: P = 2x + b

Part 2: Express the perimeter in terms of just b.

Part 2a: Express x in terms of h and b.

Answer: x = sqrt(h2+(b/2)2), from the Pythagorean theorem.

Part 2b: Express h in terms of b, and thereby x in terms of just b.

Answer: Since the area is 1, we have (1/2)bh=1, so h=2/b

We can substitute 2/b in place of h in the answer to part 2a, giving

x = sqrt((2/b)2+(b/2)2)

Answer to part 2: P = 2 sqrt((2/b)2+(b/2)2) + b

Part 3: Find the value of b for which the perimeter is minimized, and find the minimum perimeter as well.

First, find P', the derivative of P with respect to b using the exponent rule and the chain rule.  Start out by expressing P in terms of expressions with powers, to make it easier to use the exponent rule.

P = 2 (4 b-2+(1/4)b2)1/2 + b
P' = (2) (1/2) (4 b-2+(1/4)b2)-1/2 (-8b-3+(1/2)b) + 1
P' = (-8b-3+(1/2)b)/sqrt(4 b-2+(1/4)b2) + 1
P' = (-8b-3+(1/2)b+sqrt(4 b-2+(1/4)b2))/sqrt(4 b-2+(1/4)b2)

Now, set P' = 0 to find the value of b for which P is minimized (or maximized)

(-8b-3+(1/2)b+sqrt(4 b-2+(1/4)b2))/sqrt(4 b-2+(1/4)b2) = 0
-8b-3+(1/2)b+sqrt(4 b-2+(1/4)b2) = 0
8b-3-(1/2)b = sqrt(4 b-2+(1/4)b2)
64b-6-8b-2+(1/4)b2 = 4 b-2+(1/4)b2
12b-2 = 64b-6
12b4 = 64
b4 = 16/3
b = 2/31/4.

Checking in the original equation for part 3, this value of b works.  (It's important to check because we squared both sides, which has the potential to introduce spurious solutions.)

Answer to part 3: The value of b for which P' is 0 is the first answer sought in part 3:

b=2/31/4, or about 1.519671371.

When b=2/31/4, b/2=1/31/4, and h=2/b=31/4.  Square b/2 and h to get

(b/2)2 = 1/sqrt(3), h2=sqrt(3).  Add them together to get the square of x, which is

x2 = 1/sqrt(3)+sqrt(3) = 4/sqrt(3).  Take the square root of this to get

x = 2/31/4, which is the same as b.  So the minimum perimeter is 3b=6/31/4, which is the second answer sought in part 3.

As you might have expected all along, this is an equilateral triangle

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