1. ## 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.

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

2. ## Re: Webwork assistance

Originally Posted by cyrus106
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.

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.

Originally Posted by cyrus106
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

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.

3. ## Re: Webwork assistance

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