Answer:
Let the length of base of rectangle be 'x'
Let the height of rectangle be 'y'
Thus we have for rectangle
[tex]Area=Length\times Breadth\\\\Area=x\times y[/tex]
In the attached figure we can see that
[tex]tan(60^{o})=\frac{y}{\frac{(a-x)}{2}}\\\\\therefore y=\frac{\sqrt{3}}{2}(a-x)\\\\Area=x\times \frac{\sqrt{3}}{2}(a-x)\\\\[/tex]
Differentiating area with respect to x and equating the result to zero we get
[tex]A(x)=\frac{\sqrt{3}}{2}(ax-x^{2})\\\\A'(x)=\frac{\sqrt{3}}{2}(a-2x)=0\\\\\therefore x=\frac{a}{2}\\\\y=\frac{\sqrt{3}a}{4}[/tex]
