find the value of x for which the perimeter is a maximum?
The perimeter of a strange shape given by P= sqrt(2x-x^2) where x is the length of one side. Find the value of x for which the perimeter is a maximum
The perimeter of a strange shape given by P= sqrt(2x-x^2) where x is the length of one side. Find the value of x for which the perimeter is a maximum
Johan
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If the perimeter P is at its maximum, then so is its square P^2. P^2 = 2x - x^2, which is a quadratic function in x, and you probably know how to find its maximum.
Note that in the quadratic function, the coefficient of x^2 is negative, so its a "parabola on its head" and its vertex is hence a maximum. The value of x at the vertex is given by x = -b/2a, where b is the coefficient of x and a is the coefficient of x^2. So in this case:
P^2 = -x^2 + 2x + 0 ==> a = -1, b = 2, c = 0
x at vertex = -b/2a = -2/(2(-1) = 1.
So x = 1 is the answer.
By the way, the maximum perimeter is then P = sqrt(2(1)- 1^2) = sqrt(2 - 1) = sqrt(1) = 1.