Question

A normal window is constructed by adjoining a semicircle to the top of an ordinary rectangular window, (see figure ) The perimeter of the window is 12 feet. what dimensions will produce a window of maximum area? (Round you answers to two decimal places ) what is the width x= what is the length y.?
(Question2) write the function in the form f(×) = ×^3- 6×^2- 15×+9, k = -2.
f (×)=?

Answer 1

Let's find the perimeter of the window.

The bottom side is $x$. The left and right sides make $2y$.
The perimeter of a circle is $2\pi r$, so the perimeter of a semicircle must be $\pi r$, The radius is $\frac{1}2x$, so that gives $\frac{1}2\pi x$ for the curve. All of that is equal to 12.

$x+2y+\frac{1}2\pi x=12$

We only want to use one variable to create the area formula, so let's solve for $y$.

$2y=12-x-\frac{1}2\pi x$

$y=6-\frac{1}2x-\frac{1}4\pi x$

Now that we have a value for $y$ in terms of $x$, we can find the area in terms of $x$.

The area of the rectangle is going to be $xy$, which then becomes

$A_r=x(6-\frac{1}2x-\frac{1}4\pi x$

$A_r=6x-\frac{1}2x^2-\frac{1}4\pi x^2$

The area of the semicircle is going to be $\frac{1}2\pi r^2$.

Since $r=\frac{1}2x$, $A_{sc}=\frac{1}2\pi (\frac{1}2x)^2$.

$A_{sc}=\frac{1}2\pi \frac{1}4x^2$

$A_{sc}=\frac{1}8\pi x^2$

Now let's add the areas of the rectangle and semicircle.

$A=A_r+A_{sc}$

$A=6x-\frac{1}2x^2-\frac{1}4\pi x^2+\frac{1}8\pi x^2$

$A=6x-\frac{1}2x^2-\frac{1}8\pi x^2$

If you wanted to factor out $\frac{1}8$ like you did, this would become

$\boxed{A(x)=\frac{1}8(48x=4x^2-\pi x^2)}$

Now what we want to do is find what $x$ is when $A$ is at its highest point, Once we have the value for $x$ we can also find the value for $y$, of course.

Let's put our equation in the general form of a quadratic.

$A(x)=(-\frac{1}2-\frac{1}8\pi )x^2+6x$

Now we can use the vertex formula $x=\frac{-b}{2a}$.
($a$ and $b$ refer to $ax^2+bx+c$.)

$x=\frac{-6}{2(-\frac{1}2-\frac{1}8\pi)}$

$x=\frac{-6}{-\frac{1}4\pi -1}$

$x=\frac{-24}{-\pi -4}$

$\boxed{x=\frac{24}{\pi +4}}$

Now let's plug that in for $y=6-\frac{1}2x-\frac{1}4\pi x$.

Since our final answers are in decimal form and not exact form, we can make our lives a little easier here and just use $x\approx3.36059492$.

$y\approx6-\frac{1}2(3.36059492)-\frac{1}4\pi(3.36059492)$

$y\approx6-(1.68029746+2.63940507809)$

$\boxed{y\approx1.68029746191}$

Let's take our answers for $x$ and $y$ and round to 2 decimal places.

$\boxed{x\approx 3.36\ ft}$

$\boxed{y\approx 1.68\ ft}$