QUEST FOR CAMELOT. i mean, CALCULUS

[maeveoneill]

A trough is to be made by bending a long rectangular piece of tin 3 units wide. The cross section of the trough is a isosceles trapezoid with sides making angles of 120 degree with the base. Find the length of one of the sides that is bent which give a maximum capacity.


Find the area of the largest isoscles triangle ABC, AB=AC, that can be inscribed in a ellipse x²/9 + y²/4 = 1, with vertex A=(0,2).

 

[galactus]

\(\displaystyle A=\frac{1}{2}h(a+b)\)

\(\displaystyle h=xsin(\frac{{\pi}}{3})\)

\(\displaystyle a=3-2x\)

\(\displaystyle b=3-2x+2xcos(\frac{{\pi}}{3})\)

\(\displaystyle A=\frac{1}{2}(xsin(\frac{{\pi}}{3}))(6-4x+2xcos(\frac{{\pi}}{3}))\)

\(\displaystyle =\frac{-3sqrt{3}x(x-2)}{4}\)

Now take the derivative:

\(\displaystyle \frac{-3\sqrt{3}(x-1)}{2}\)

Setting to 0 and solving for x we can see x=1.

Therefore, max area is attained when all three sides have the same length of 1 unit.

This is the answer to be expected. Also, it can be shown that the max area is obtained when the inside angle with the base is 120 degrees(what they gave you). Therefore, the sides will have equal length.

 

[soroban]

Hello, maeveoneill!

Find the area of the largest isoscles triangle \(\displaystyle ABC,\;AB\,=\,AC\),
that can be inscribed in a ellipse \(\displaystyle \L\frac{x^2}{9}\,+\,\frac{y^2}{4}\:=\:1\) with vertex \(\displaystyle A(0,2)\).

               A|(0,2)
              * * *
           *   /|\   *
          *   / | \   *
    - - - * -/- + -\- * + -
          * /   |   \ *
          B*----+----*C
      (-x,y)  * * *  (x,y)
                |

Solve the equation for \(\displaystyle y\):

Multiply by 36: \(\displaystyle \:4x^2\,+\,9y^2\:=\:36\)

\(\displaystyle \;\;\;9y^2\:=\:36\,-\,4x^2\:=\:4(9\,-\,x^2)\;\;\Rightarrow\;\;y^2\:=\:\frac{4}{9}(9\,-\,x^2)\;\;\Rightarrow\;\;y\:=\:\pm\frac{2}{3}\sqrt{9 - x^2}\)

Since we are concerned with the lower half: \(\displaystyle y\;=\;-\frac{2}{3}\sqrt{9\,-\,x^2}\;\) [1]


The base of the triangle is: \(\displaystyle BC\,=\,2x\)

The height of the triangle is: \(\displaystyle 2\,-\,y\)

The area of the triangle is: \(\displaystyle \,A\:=\:\frac{1}{2}(2x)(2\,-\,y)\:=\:x(2\,-\,y)\)

Substitute [1]: \(\displaystyle \;A\:=\:x\left(2\,+\,\frac{2}{3}\sqrt{9\,-\,x^2}\right)\)

\(\displaystyle \;\;\)And that is the function we must maximize . . .