Thx2xr=Vπ1/3and soA= 4πVπ2/3.Answer:A(V) = 4πVπ2/3,0< V <+∞.It is permissible, and sometimes desirable, to use instead the domain 0V <+∞.C01S0M.017:The following figure shows an equilateral triangle with sides of length 2xand an altitudeof lengthh. BecauseTis a right triangle, we see thatx2+h2= (2x)2,so thath=x√3.The area of this triangle isA=hxand its perimeter isP= 6x. SoA=x2√3andx=P6.ThereforeA(P) =P2√336,0< P <∞.C01S0M.018:The square has perimeterxand thus edge lengthy=14x. The circle has circumference100−x. Thus ifzis the radius of the circle, then 2πz= 100−x, so thatz= (100−x)/(2π). The area ofthe square isy2and the area of the circle isπz2, so that the sum of the areas of the square and the circle is
This is the end of the preview.
access the rest of the document.