A sector with a particular fixed area has radius x cm - Scottish Highers Maths - Question 9 - 2018

To find the critical points, we differentiate P:

$P' = 2 - \frac{128}{x^2}$

Set the derivative equal to zero to find the values of x that yield critical points:

$2 - \frac{128}{x^2} = 0 \implies \frac{128}{x^2} = 2 \implies x^2 = 64 \implies x = 8$

To confirm that this critical point is a minimum, we check the sign of the derivative:

• For x < 8, P' is positive (increasing).
• For x > 8, P' is negative (decreasing).

Thus, x = 8 is a minimum point.

Now we substitute x = 8 back into the equation for P:

$P(8) = 2(8) + \frac{128}{8} = 16 + 16 = 32$

Therefore, the minimum value of P is 32.