SOLUTION
From the diagram below
From the diagram, we have the hypotenuse of triagle ABD as 1 + 1 = 2cm
So using SOHCAHTOA, we have
[tex]\begin{gathered} sin\varnothing=\frac{oppsite}{hypotenuse} \\ sin\varnothing=\frac{x}{2} \\ x=2sin\varnothing \\ Also\text{ } \\ cos\varnothing=\frac{adjacent}{hypotenuse} \\ cos\varnothing=\frac{y}{2} \\ y=2cos\varnothing \end{gathered}[/tex]
hence the area of the inscribed rectangle becomes
[tex]\begin{gathered} Area=x\times y \\ Area=(2sin\varnothing)(2cos\varnothing) \\ =4sin\varnothing cos\varnothing \end{gathered}[/tex]
Hence the answer is
[tex]Area=4sin\varnothing cos\varnothing[/tex]
(b) To get the maximum area, we find the derivative of the function above, we have
[tex]\begin{gathered} A=4sin\varnothing cos\varnothing \\ simplify,\text{ we have } \\ 2sin(2\varnothing) \\ =\frac{d}{d\varnothing}(2sin(2\varnothing)) \\ take\text{ the constant out } \\ =2\frac{d}{d\varnothing}(sin(2\varnothing)) \end{gathered}[/tex]
Apply the chain rule
[tex]\begin{gathered} cos(2\varnothing)\frac{d}{d\varnothing}(2\varnothing) \\ =\frac{d}{d\varnothing}(2\varnothing)=2 \\ =2cos(2\varnothing)\times2 \\ =4cos(2\varnothing) \end{gathered}[/tex]
At maximum the derivative is 0, equating to zero, we have
[tex]\begin{gathered} 4cos(2\varnothing)=0 \\ \varnothing=45\degree \end{gathered}[/tex]
So, we put this 45 into the expression for area, we have
[tex]\begin{gathered} Area=4sin\varnothing cos\varnothing \\ =4sin45\degree cos45\degree \\ =4\times\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2} \\ =\frac{4\times2}{4} \\ =2units^2 \end{gathered}[/tex]
Hence the answer is 2 square units
