Solving Optimization Problems

Key Questions

  • How do you find the dimensions of a rectangle whose area is 100 square meters and whose perimeter is a minimum?

    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(1-100/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 minimum

    I hope that this was helpful.

    Hence, the dimensions are #10\times10#.

    Wataru · 7 · Oct 8 2014
  • How do you find the dimensions of the rectangle with largest area that can be inscribed in a semicircle of radius #r# ?

    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^2-x^2#

    #y=sqrt(r^2-x^2)#

    Plugging this value in equation #(2)#

    #A=2xsqrt(r^2-x^2)#

    Differentiating wrt #x# using the product rule

    #(dA)/dx=2sqrt(r^2-x^2)-2x^2/sqrt(r^2-x^2)#

    #=(2r^2-2x^2-2x^2)/(sqrt(r^2-x^2))#

    #=(2r^2-4x^2)/(sqrt(r^2-x^2))#

    The critical points are when

    #(dA)/dx=0#

    That is

    #(2r^2-4x^2)/(sqrt(r^2-x^2))=0#

    #r^2=2x^2#

    #x=r/sqrt2#

    Then,

    #y=sqrt(r^2-x^2)=sqrt(r^2-r^2/2)=r/sqrt2#

    The maximum area is

    #A=2*r/sqrt2*r/sqrt2=r^2#

    Narad T. · 1 · Aug 12 2018
  • How do you find the points on the ellipse #4x^2+y^2=4# that are farthest from the point #(1,0)#?

    Let #(x,y)# be a point on the ellipse #4x^2+y^2=4#.

    #Leftrightarrow y^2=4-4x^2 Leftrightarrow y=pm2sqrt{1-x^2}#

    The distance #d(x)# between #(x,y)# and #(1,0)# can be expressed as

    #d(x)=sqrt{(x-1)^2+y^2}#

    by #y^2=4-4x^2#,

    #=sqrt{(x-1)^2+4-4x^2}#

    by multiplying out

    #=sqrt{-3x^2-2x+5}#

    Let us maximize #f(x)=-3x^2-2x+5#

    #f'(x)=-6x-2=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)#.

    Wataru · 30 · Sep 19 2014

Questions

Applications of Derivatives