First of all, it is easy to notice that the triangle is a right angle triangle because it has a base which is the diameter of the circle. that means that BC^2=AC^2 + AB^2 we can add and subtract one AC which will not change the equation. BC^2=AC^2 + AB^2 -AC + AC . Let us write BC^2 as (2r)^2 which will be 4r^2 and let's group the right part properly. 4r^2= (AB^2+AC)+AC^2-AC . =>
AB^2+AC= 4r^2 -AC^2+AC which can be considered as a function f(AC) where AC is the unknown variable. We need to find the maximum value of function f(AC) which can be done with differentiation . f '(AC)= 0 -2*AC +1 .
-2*AC+1=0 => AC=1/2 => we can easily notice that the function increases up to f(1/2) and then it decreases so the maximum value is at AC=1/2 which gives the full answer of 4*r*r -1/4+1/2 =4*r*r+1/4.
AB^2+AC= 4r^2 -AC^2+AC which can be considered as a function f(AC) where AC is the unknown variable. We need to find the maximum value of function f(AC) which can be done with differentiation . f '(AC)= 0 -2*AC +1 .
-2*AC+1=0 => AC=1/2 => we can easily notice that the function increases up to f(1/2) and then it decreases so the maximum value is at AC=1/2 which gives the full answer of 4*r*r -1/4+1/2 =4*r*r+1/4.
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