Maximum Area

[Mr.Pacman]

So i got to this problem in my math book and wanted to try it but i got stuck
I know to find maximum area you should take the first derivative set it equal to zero
and so on...but the fact that its bounded in side the triangle is throwing me off...


Problem:

The Figues show a rectange, a circle, and a semicircle inscribed in a triangle bounded by the coordinate axes and the first-quadrant portion of the line with intercepts (3,0) and (0,4). Find the dimensions of each inscribed figure such that its area is maximum. State whether calculus was helpful in finding the required dimensions. Explain your reasoning....

 

[galactus]

Here's the first one. A freebie.

We can use similar triangles.

The area of the rectangle is \(\displaystyle A=xy\)

Let x and y be the dimensions of the rectangle.

Then, by similar triangles, we have \(\displaystyle \frac{x}{3}=\frac{4-y}{4}\)

\(\displaystyle y=4-\frac{4x}{3}\)

\(\displaystyle A=x\left(4-\frac{4x}{3}\right)\)

Differentiate:

\(\displaystyle \frac{dA}{dx}=4-\frac{8x}{3}\)

Setting to 0 and solving for x gives us \(\displaystyle x=\frac{3}{2}\)

and \(\displaystyle y=2\)

The max area is \(\displaystyle 2(\frac{3}{2})=3\)


EDIT: I got mixed up and thought of the centroid. Soroban straightened out my mistake.

 

[Mr.Pacman]

Thanks but theres one part im not understanding



Then, by similar triangles, we have \(\displaystyle \frac{x}{3}=\frac{4-y}{4}\)

\(\displaystyle y=4-\frac{4x}{3}\)


where did you pull the numbers from for the rectangle
because if the \(\displaystyle A=lw\) thus \(\displaystyle A=(1)(2)\)

And for the for the circle

A=pie * r[sup:vcjadcvt]2[/sup:vcjadcvt]


with pie being a constant
i can differentiate for r correct.

and then plug in 1 for the radius, since the r's length is from (0,0) to point (1,0)
then set it equal to zero to find the max?

 

[soroban]

Hello, Mr.Pacman!

The second and third parts do not require Calculus . . . just geometry.

The figues show a rectange, a circle, and a semicircle inscribed in a 3-4-5 right triangle.
Find the dimensions of each inscribed figure such that its area is maximum.

\(\displaystyle \text{Second problem: an inscribed circle.}\)
. . \(\displaystyle \text{Important: there is only }one\text{ such circle.}\)

\(\displaystyle \text{We see that the center of the circle is at }(r,r)\,\text{ and the radius is }r.\)
[The incenter is the intersection of the angle bisectors . . . not the centroid.]

\(\displaystyle \text{The center is }r \text{ units from the hypotenuse: }\:4x + 3y - 12 \:=\:0\)

. . \(\displaystyle \text{That is: }\;\frac{|4r + 3r - 12|}{\sqrt{4^2+3^2}} \:=\:r \quad\Rightarrow\quad |7r-12| \:=\:5r \quad\Rightarrow\quad r \:=\:1,6\)

\(\displaystyle \text{Since }r \,=\,1\text{, the area is: }\:A \:=\:\pi(1^2) \:=\:\pi\text{ units}^2.\)



\(\displaystyle \text{Third problem: an inscribed semicircle.}\)
. . \(\displaystyle \text{Again, there is }one\text{ such semicircle.}\)

\(\displaystyle \text{The center is on the hypotenuse; draw perpendiculars to the }x\text{- and }y\text{-axes.}\)
. . \(\displaystyle \text{Their lengths are both }r\text{, the radius of the semiclrcle}\)

\(\displaystyle \text{We see that the center is }(r,r)\text{ and it lies on the line: }\:4x + 3y \:=\:12\)
. . \(\displaystyle 4r + 3r \:=\:12 \quad\Rightarrow\quad r \:=\:\tfrac{12}{7}\)

\(\displaystyle \text{Therefore, the area is: }\;A \:=\:\frac{1}{2}\pi\left(\frac{12}{7}\right)^2 \:=\:\frac{72\pi}{49}\text{ units}^2.\)

 

[galactus]

Yep, I got mixed up and thought of the centroid. Thanks for that correction Soroban.

 

[soroban]

Hello, galactus!

I used to get those centers confused.
Finally, I used baby-talk to sort them out in my mind.


Circumcircle (circumscribed circle)
. . Center is equidistant from the vertices:
. . . . intersection of the perpendicular bisectors.

Incircle (inscribed circle)
. . Center is equidistant from the sides:
. . . . intersection of the angle bisectors.

Centroid: center of mass
. . This "bisects the area":
. . . . centroid is the intersection of the medians.