Solving Optimization Problems
Key Questions

Let
#x# and#y# be the base and the height of the rectangle, respectively.
Since the area is 100
#m^2# ,#xy=100 Rightarrow y=100/x# The perimeter
#P# can be expressed as#P=2(x+y)=2(x+100/x)# So, we want to minimize
#P(x)# on#(0,infty)# .By taking the derivative,
#P'(x)=2(1100/x^2)=0 Rightarrowx=pm10# #x=10# is the only critical value on#(0,infty)# #y=100/10=10# By testing some sample values,
#P'(1)<0 Rightarrow P(x)# is decreasing on#(0,10]# .#P'(11)>0 Rightarrow P(x)# is increasing on#[10,infty)# Therefore,
#P(10)# is the minimumI hope that this was helpful.
Hence, the dimensions are
#10\times10# . 
Answer:
The dimensions of the rectangle is
#sqrt2r# and#r/sqrt2# Explanation:
The equation of the semicircle is
#x^2+y^2=r^2# .......................#(1)# The area of the rectangle is
#A=2xy# ....................#(2)# From equation
#(1)# , we get#y^2=r^2x^2# #y=sqrt(r^2x^2)# Plugging this value in equation
#(2)# #A=2xsqrt(r^2x^2)# Differentiating wrt
#x# using the product rule#(dA)/dx=2sqrt(r^2x^2)2x^2/sqrt(r^2x^2)# #=(2r^22x^22x^2)/(sqrt(r^2x^2))# #=(2r^24x^2)/(sqrt(r^2x^2))# The critical points are when
#(dA)/dx=0# That is
#(2r^24x^2)/(sqrt(r^2x^2))=0# #r^2=2x^2# #x=r/sqrt2# Then,
#y=sqrt(r^2x^2)=sqrt(r^2r^2/2)=r/sqrt2# The maximum area is
#A=2*r/sqrt2*r/sqrt2=r^2# 
Let
#(x,y)# be a point on the ellipse#4x^2+y^2=4# .#Leftrightarrow y^2=44x^2 Leftrightarrow y=pm2sqrt{1x^2}# The distance
#d(x)# between#(x,y)# and#(1,0)# can be expressed as#d(x)=sqrt{(x1)^2+y^2}# by
#y^2=44x^2# ,#=sqrt{(x1)^2+44x^2}# by multiplying out
#=sqrt{3x^22x+5}# Let us maximize
#f(x)=3x^22x+5# #f'(x)=6x2=0 Rightarrow x=1/3# (the only critical value)#f''(x)=6 Rightarrow x=1/3# maximizes#f(x)# and#d(x)# Since
#y=pm2sqrt{1(1/3)^2}=pm{4sqrt{2}}/3# ,the farthest points are
#(1/3,pm{4sqrt{2}}/3)# .