
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(h^{2}+(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)b^{2})^{1/2} + b
P' = (2) (1/2) (4 b^{2}+(1/4)b^{2})^{1/2} (8b^{3}+(1/2)b) + 1
P' = (8b^{3}+(1/2)b)/sqrt(4 b^{2}+(1/4)b^{2}) + 1
P' = (8b^{3}+(1/2)b+sqrt(4 b^{2}+(1/4)b^{2}))/sqrt(4 b^{2}+(1/4)b^{2})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)b^{2}))/sqrt(4 b^{2}+(1/4)b^{2}) = 0
8b^{3}+(1/2)b+sqrt(4 b^{2}+(1/4)b^{2}) = 0
8b^{3}(1/2)b = sqrt(4 b^{2}+(1/4)b^{2})
64b^{6}8b^{2}+(1/4)b^{2} = 4 b^{2}+(1/4)b^{2}
12b^{2} = 64b^{6}
12b^{4} = 64
b^{4} = 16/3
b = 2/3^{1/4}.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=2/3^{1/4}, or about 1.519671371.
When b=2/3^{1/4}, b/2=1/3^{1/4}, and h=2/b=3^{1/4}. Square b/2 and h to get
(b/2)^{2} = 1/sqrt(3), h^{2}=sqrt(3). Add them together to get the square of x, which is
x^{2} = 1/sqrt(3)+sqrt(3) = 4/sqrt(3). Take the square root of this to get
x = 2/3^{1/4}, which is the same as b. So the minimum perimeter is 3b=6/3^{1/4}, 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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