Results 1 to 10 of 10

Math Help - max/min

  1. #1
    Junior Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    44

    max/min

    A rectangular tank with a bottom and sided but but no top is to have volume 500 cubic feet. Determine the

    dimensions (length, width, height) with the smallest possible surface area.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: max/min

    Quote Originally Posted by apatite View Post
    A rectangular tank with a bottom and sided but but no top is to have volume 500 cubic feet. Determine the

    dimensions (length, width, height) with the smallest possible surface area.
    Hi apatite!

    For a minimal surface area, the top of the tank will be square.
    Let's say the side of this square is x.
    And let h be the height of the tank.

    Then the volume is x \times x \times h = 500 and the surface area is A = x \times x + 4 x \times h.

    Can you combine these equations to find A as a function of x?
    Afterwards its derivative needs to be zero, which will give you x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    44

    Re: max/min

    Quote Originally Posted by ILikeSerena View Post
    Hi apatite!

    For a minimal surface area, the top of the tank will be square.
    Let's say the side of this square is x.
    And let h be the height of the tank.

    Then the volume is x \times x \times h = 500 and the surface area is A = x \times x + 4 x \times h.

    Can you combine these equations to find A as a function of x?
    Afterwards its derivative needs to be zero, which will give you x.
    derivative becomes zero. I think your answer is wrong bud
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: max/min

    Quote Originally Posted by apatite View Post
    derivative becomes zero. I think your answer is wrong bud
    Neh, it only means you did not combine the equations as I suggested... bud!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    44

    Re: max/min

    Quote Originally Posted by ILikeSerena View Post
    Neh, it only means you did not combine the equations as I suggested... bud!
    what do you mean? h= 500/x^2 and then you substitute that in the second equation and set derivative equal zero?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: max/min

    Quote Originally Posted by apatite View Post
    what do you mean? h= 500/x^2 and then you substitute that in the second equation and set derivative equal zero?
    Yep!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    44

    Re: max/min

    Quote Originally Posted by ILikeSerena View Post
    Yep!
    A= 5x^2 * 500/x^2 the x^2 cancels out. That's what I don't get.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: max/min

    Quote Originally Posted by ILikeSerena View Post
    Hi apatite!

    For a minimal surface area, the top of the tank will be square.
    Let's say the side of this square is x.
    And let h be the height of the tank.

    Then the volume is x \times x \times h = 500 and the surface area is A = x \times x + 4 x \times h.

    Can you combine these equations to find A as a function of x?
    Afterwards its derivative needs to be zero, which will give you x.
    Quote Originally Posted by apatite View Post
    A= 5x^2 * 500/x^2 the x^2 cancels out. That's what I don't get.
    Looks like you did not substitute correctly.

    x \times x \times h = 500

    h = \frac {500}{x \times x} \qquad (1)

    Substitute (1) into the formula for the area:

    A = x \times x + 4 x \times h

    A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    44

    Re: max/min

    Quote Originally Posted by ILikeSerena View Post
    Looks like you did not substitute correctly.

    x \times x \times h = 500

    h = \frac {500}{x \times x} \qquad (1)

    Substitute (1) into the formula for the area:

    A = x \times x + 4 x \times h

    A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)
    ooo so A= (x^2+4x) * (5oo/x^2). Thank you for your help
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: max/min

    Quote Originally Posted by ILikeSerena View Post
    Looks like you did not substitute correctly.
    x \times x \times h = 500

    h = \frac {500}{x \times x} \qquad (1)

    Substitute (1) into the formula for the area:
    A = x \times x + 4 x \times h

    A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)
    Quote Originally Posted by apatite View Post
    ooo so A= (x^2+4x) * (5oo/x^2). Thank you for your help
    Not quite.
    Multiplication has a higher priority than addition.
    A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)

    A = x \times x +  \left(\frac {4 x \times 500}{x \times x}\right)

    A = x^2 +  \frac {2000}{x}
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum