Find the lenght of the sides of an isosceles triangle with a given perimeter if its area is to be as great as possible.
let k be the the length of the two iso sides, so the other side is P-2k. Call that the base.
Using Heron's formula:
Area=SQRT(s(s-a)(s-b)(s-c)),
but s= P/2 , a and b are equal, and c is P-2a
area= sqrt(s(s-a)^2 (s-P+2a))
dA/d=0= you do it. Solve for a
To find the length of the sides of an isosceles triangle with a given perimeter, where the area is as great as possible, we can use calculus to maximize the area function.
First, let's assume that the perimeter of the triangle is represented by P. Since it is an isosceles triangle, let k be the length of the two equal sides, and thus the other side will be P - 2k.
Using Heron's formula to calculate the area of the triangle:
Area = √[s(s - a)(s - b)(s - c)]
where s is the semi-perimeter (s = P/2), a and b are the equal sides (length k), and c is the base (P - 2k).
So, the area can be rewritten as:
Area = √[s(s - k)(s - k)(s - (P - 2k))]
Now, let's derive the area function with respect to k and set it equal to 0 to find the maximum area:
dA/dk = 0
To start, we need to simplify the expression √[s(s - k)(s - k)(s - (P - 2k))] before taking the derivative.
1. Expand (s - (P - 2k)):
s - (P - 2k) = s - P + 2k
2. Substitute this expression back into the area formula:
Area = √[s(s - k)(s - k)(s - P + 2k)]
3. Multiply the terms inside the square root:
Area = √[s(s - k)^2(s - P + 2k)]
Now, let's differentiate the area function with respect to k:
dA/dk = 0 = d/dk[√[s(s - k)^2(s - P + 2k)]]
To simplify the differentiation, we can apply the chain rule:
dA/dk = 0 = (1/2) [s(s - k)^2(s - P + 2k)]^(-1/2) * d/dk[s(s - k)^2(s - P + 2k)]
4. Calculate the derivative term by term:
dA/dk = (1/2) [s(s - k)^2(s - P + 2k)]^(-1/2) * [s(2(s - k)(-1)) + 2(s - P + 2k)(1)]
5. Simplify the equation:
(1/2) [s(s - k)^2(s - P + 2k)]^(-1/2) * [2s(s - k) - 2(s - P + 2k)]
6. Further simplify the equation:
0 = [2s(s - k) - 2(s - P + 2k)] / 2[s(s - k)^2(s - P + 2k)]^(1/2)
7. Cancel out the common factors:
0 = [s(s - k) - (s - P + 2k)] / [s(s - k)^2(s - P + 2k)]^(1/2)
8. Multiply both sides by the denominator to eliminate the square root:
0 = s(s - k) - (s - P + 2k)
0 = s^2 - sk - s + P - 2k
9. Rearrange the terms:
0 = s^2 - s - sk - 2k + P
10. Combine like terms:
0 = s^2 - s(1 + k) - k(2 - P)
11. Use the quadratic formula to solve for s:
s = [-(-1 - k) ± √((1 + k)^2 - 4(2 - P)k)] / 2
12. Simplify and solve for s:
s = (1 + k ± √[1 + 2k + k^2 - 8k + 4Pk]) / 2
s = (1 + k ± √[k^2 - 6k + P(4 - k)]) / 2
By solving the equation for s, we can find the length of the equal sides of the isosceles triangle that maximizes the area, given the perimeter P.