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Thread: Webwork assistance

  1. #1
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    Webwork assistance

    I have solved a few math problems for my webwork homework, yet when I enter them into the website it tells me my answers are wrong and I am becoming quite frustrated, can anybody help me?

    Ex:
    If 1100 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

    my answer:
    3263 or 3262, depending on how you round it upboth entered in as attempts, both wrong)


    Ex2: A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 46 feet?

    (pi x r) + 6r

    Now, set that equal to 46 feet and solve for r (factor it out in the equation)

    r(pi + 6) = 46
    9.14 r = 46
    r = (around) 5.03 feet

    Now, the AREA of that window is half the area of a circle of radius r, PLUS the area of the supporting rectangle, or the length of the sides (2r) squared.

    1/2 (pi x r^2) + 10.06^2

    my answer: 141 square feet(wrong)
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  2. #2
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    Re: Webwork assistance

    Quote Originally Posted by cyrus106 View Post
    I have solved a few math problems for my webwork homework, yet when I enter them into the website it tells me my answers are wrong and I am becoming quite frustrated, can anybody help me?

    Ex:
    If 1100 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

    my answer:
    3263 or 3262, depending on how you round it upboth entered in as attempts, both wrong)
    The dimensions of the box could be $\displaystyle x \times x \times y$. Then, the box has four sides with dimension $\displaystyle x \times y$ and one side with dimension $\displaystyle x\times x$. Hence, $\displaystyle x^2 + 4xy = 1100$. You want to maximize the volume, which is $\displaystyle V(x,y) = x^2y$. Now, from the first equation, you should solve for $\displaystyle y$: $\displaystyle y = \dfrac{1100-x^2}{4x} = 275x^{-1} - \dfrac{1}{4}x$. Plugging that in to the volume formula:

    $\displaystyle V(x) = x^2\left(275x^{-1} - \dfrac{1}{4}x\right) = 275x - \dfrac{1}{4}x^3$

    Now, to maximize volume, take the derivative and set it equal to zero:

    $\displaystyle V'(x) = 275 - \dfrac{3}{4}x^2 = 0$. So, $\displaystyle x = \pm \sqrt{\dfrac{1100}{3}}$. Check the second derivative: $\displaystyle V''(x) = -\dfrac{3}{2}x$. Since $\displaystyle V''\left(\sqrt{\dfrac{1100}{3}}\right) < 0$, there is a local maximum at $\displaystyle x = \sqrt{\dfrac{1100}{3}} = \dfrac{10}{3}\sqrt{33}$. Hence, the maximum area is given by:

    $\displaystyle V\left(\dfrac{10}{3}\sqrt{33}\right) = 275\dfrac{10}{3}\sqrt{33} - \dfrac{1}{4}\dfrac{10^3}{3^3}\sqrt{33^3} \approx 3510.566$ so $\displaystyle 3511$ should be the correct answer.

    Quote Originally Posted by cyrus106 View Post
    Ex2: A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 46 feet?

    (pi x r) + 6r

    Now, set that equal to 46 feet and solve for r (factor it out in the equation)

    r(pi + 6) = 46
    9.14 r = 46
    r = (around) 5.03 feet

    Now, the AREA of that window is half the area of a circle of radius r, PLUS the area of the supporting rectangle, or the length of the sides (2r) squared.

    1/2 (pi x r^2) + 10.06^2

    my answer: 141 square feet(wrong)
    First, I would recommend rounding at the very end. Second, the width of the window is $\displaystyle 2r$, as you suggested. But, the height of the rectangular portion could be anything. So, let the height of the window be $\displaystyle h$.

    $\displaystyle \pi r + 2h + 2r = 46$

    $\displaystyle h = \dfrac{46-2r-\pi r}{2}$

    Then, area is:

    $\displaystyle A(r,h) = \dfrac{1}{2}\pi r^2 + 2rh$

    Plugging in for $\displaystyle h$, you have:

    $\displaystyle \begin{align*}A(r) & = \dfrac{1}{2}\pi r^2 + 2r\dfrac{46 - 2r - \pi r}{2} \\ & = \dfrac{1}{2}\pi r^2 +46r - 2r^2 - \pi r^2 \\ & = 46r - \dfrac{4+\pi}{2}r^2\end{align*}$

    Then $\displaystyle A'(r) = 46 - (4+\pi)r$. Setting it equal to zero, you find $\displaystyle r = \dfrac{46}{4+\pi}$. The second derivative is always negative, so the area of the window is maximized at that value. So, the maximum area is $\displaystyle A\left(\dfrac{46}{4+\pi}\right) = \dfrac{46^2}{4+\pi} - \dfrac{4+\pi}{2}\cdot \dfrac{46^2}{(4+\pi)^2} = \dfrac{46^2}{2(4+\pi)} \approx 148.15$, so 148 square feet should be the correct answer.
    Last edited by SlipEternal; Nov 13th 2013 at 06:38 AM.
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  3. #3
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    Re: Webwork assistance

    Wow! Much thanks my friend! Your explanations were spot on, I'll be sure to remember these tips
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