The problem is the following (Ron Larson's Calculus for AP - 2nd Edition):
My solution to part a). was found by proving that the triangles are similar by the AA property (by comparing Alternate Interior Angles) and that they have a side ratio of "k". My answer to part a). is:
$$ A(x)=\frac{1}{2}dx (1-k)+\frac{1}{2}kdb. $$
Part b) and c) however seem to indicate that there is a maximum and minimum function, but since the function is a line, there shouldn't be one. Now I am beginning to doubt the correctness of my solution for part a).
If a). is correct, then how do I proceed? If a). is incorrect, what mistake have I made?
Your answer to part a) is not correct. The function $A(x)$ should be equal to $$\frac{dx}{2}+\frac{D(b-x)}{2}$$ where $D$ can be obtained by the similarity of the two triangles $$k(x)=\frac{D}{d}=\frac{b-x}{x}.$$ Hence, for $x\in (0,b]$, we have $$A(x)=\frac{dx}{2}+\frac{d(b-x)^2}{2x}$$ which is not linear. Taking the derivative we find $$A'(x)=\frac{d}{2}\left(1+\frac{-2(b-x)x-(b-x)^2}{x^2}\right)=\frac{d(2x^2-b^2)}{2x^2}.$$ Hence, in $(0,b]$, $A$ has no the maximum value because $A(x)$ tends to $+\infty$ as $x\to 0^+$, and it attains the minimum value for $x=\frac{b}{\sqrt{2}}$, that is when $|QS|=d(\sqrt{2}-1)$.