First, we set that #x# is the width and #y# is the height.
Also, we note that the outer perimeter of the semicircle is the perimeter of a circle with radius #x# divided by 2.
"The perimeter is 30ft."
This means that:#x+2y+(pix)/2=30#
The area of the window would be the rectangle plus the half-circle.
So we have: #xy+(pi(x/2)^2)/2=A#
Let's manipulate the first equation so that we have #y# in terms of #x#.
#x+2y+(pix)/2=30#
#=>x+(pix)/2=30-2y#
#=>x+(pix)/2-30=-2y#
#=>-x/2-(pix)/4+15=y#
We plug this in the second equation.
#xy+(pi(x/2)^2)/2=A=>x(-x/2-(pix)/4+15)+(pi(x/2)^2)/2=A# We simplify this function.
#x(-x/2-(pix)/4+15)+(pi(x/2)^2)/2=A#
#=>x(-(2x)/4-(pix)/4+15)+pi(x/2)^2*1/2=A#
#=>x(-(2x)/4-(pix)/4+15)+pi*(x^2)/4*1/2=A#
#=>x(-1((2x+pix)/4)+15)+pi*(x^2)/4*1/2=A#
#=>x(-1((x(2+pi))/4)+15)+pi*(x^2)/4*1/2=A#
#=>x(-1((x(2+pi))/4)+15)+(pix^2)/8=A#
#=>-x((x(2+pi))/4)+15x+(pix^2)/8=A#
#=>(-x*x(2+pi))/4+15x+(pix^2)/8=A#
#=>(-x^2(2+pi))/4+15x+(pix^2)/8=A#
#=>(-2(x^2(2+pi)))/8+15x+(pix^2)/8=A#
#=>(-2x^2(2+pi))/8+(pix^2)/8+15x=A#
#=>(-2x^2(2+pi)+pix^2)/8+15x=A#
#=>(-x^2[2(2+pi)-pi])/8+(120x)/8=A#
#=>(-x^2[4+2pi-pi])/8+(120x)/8=A#
#=>(-x^2[4+pi]+120x)/8=A#
This is the function that relates #x# to the total area of the window.
Let's turn this equation around a bit.
#(-x^2[4+pi]+120x)/8=A=>-x^2[4+pi]*1/8+15x=A#
This is a quadratic function in the form of #ax^2+bx+c=f(x)#
#a=-[4+pi]/8#
#b=15#
#c=0#
Since #a# is a negative number, this function has a maximum point#(h,k)#.
Remember that #h=(-b)/(2a)#
Therefore, #h=(-15)/(2(-[4+pi]/8))#
#h=(-15)/(-[4+pi]/4)#
#h=-15*4/-[4+pi]#
#h=60/[4+pi]# which is approximately #8.401#.
We plug this into our original function to get the value of #k#.
#-(8.401)^2[4+pi]*1/8+15(8.401)~~k#
#k# is approximately #63.011#
This means that the highest area possible of the window is around #63.011ft^2#
when #x~~8.401ft#
Using the fact that #x+2y+(pix)/2=30#, we can solve for #y#.
#8.401+2y+(pi*8.401)/2=30#
#=>8.401+2y=30-(pi*8.401)/2#
#=>2y=30-(pi*8.401)/2-8.401#
#=>y=(30-(pi*8.401)/2-8.401)/2#
#y# is approximately #4.201#
That is our answer! (I have to admit, this is a very painful problem to solve)
