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 xx and yy be the base and the height of the rectangle, respectively.

    Since the area is 100 m^2m2,

    xy=100 Rightarrow y=100/xxy=100⇒y=100x

    The perimeter PP can be expressed as

    P=2(x+y)=2(x+100/x)P=2(x+y)=2(x+100x)

    So, we want to minimize P(x)P(x) on (0,infty)(0,∞).

    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