Optimization: largest rectangle, area in wire, boat distance

[MarkSA]

I'm pretty lost on these. Wonder if someone can give me some nudges in the right direction.

1) Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y = 8 - x^2

For this one I have the picture drawn.. thats about it. Not sure where to begin.

2) A piece of wire 10m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is a) a minimum b) a maximum?

Oh dear...

3) A boat leaves a dock at 2:00 pm and travels due south at a speed of 20km/h. Another boat has been heading due east at 15km/hr and reaches the same dock at 3:00PM. At what time were the two boats closest together?)

It looks like a right triangle might be used in this one, and since it's time, i'm guessing I need to work t=d/r into it. Once again i'm stuck on where to begin this though.

 

[soroban]

Re: Optimization

Hello, "MarkSA!

3) A boat leaves a dock at 2:00 pm and travels due south at a speed of 20 km/h.
Another boat has been heading due east at 15 km/hr and reaches the same dock at 3:00 pm.
At what time were the two boats closest together?

. . : - - - - - 15  - - - - - :
    P    15t    B    15-15t   D
    *-----------*-------------*
                  *           |
                    *         |
                   x  *       | 20t
                        *     |
                          *   |
                            * |
                              * A

At 2:00 the first boat starts at \(\displaystyle D\) and travels south at 20 kph.
. . In \(\displaystyle t\) hours, it has moved \(\displaystyle 20t\) km to point \(\displaystyle A.\)

The second boat starts at \(\displaystyle P\) and moves east at 15 kph and reaches \(\displaystyle D\) at 3:00.
. . Hence, at 2:00 it was 15 km west of \(\displaystyle D\): \(\displaystyle \,PD \,=\,15\)
In \(\displaystyle t\) hours, it has moved \(\displaystyle 15t\) km to point \(\displaystyle B:\;BD\,=\,15\,-\,15t\)

We wish to minimize the distance \(\displaystyle x\,=\,AB\)
. . where: \(\displaystyle \:x\:=\:\sqrt{(20t)^2\,+\,(15\,-\,15t)^2}\)

We have: \(\displaystyle \:x\:=\:\left(625t^2\,-\,450t\,+\,225\right)^{\frac{1}{2}}\)

. . Go for it!

 

[soroban]

Re: Optimization

Hello, MarkSA!

1) Find the dimensions of the rectangle of largest area that has its base on the x-axis
and its other two vertices above the x-axis and lying on the parabola: \(\displaystyle \,y \:=\: 8\, -\, x^2\)

                    |
                    *
                *   |   *  (x,y)
             *------+------*
           * |      |      | *
          *  |      |      |  *
             |      |     y|
         *   |      |      |   *
             |      |      |
             |      |      |
      - * - -+------+------+- - * -
                    |   x

The length of the rectangle is: \(\displaystyle 2x.\)
The height is: \(\displaystyle y \:=\:8\,-\,x^2\)

The area of the rectangle is: \(\displaystyle \;A \:=\:2x(8\,-\,x^2) \:=\:16x\,-\,2x^3\)

. . Now maximize it . . .

 

[galactus]

Re: Optimization

2) A piece of wire 10m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is a) a minimum b) a maximum?

Oh dear...

For the minimum:

Let the section of the wire which comprises the square be 's' and the length for the triangle be 't'.

The total length of the wire is then s+t=10......[1]

Then, one side of the square would have length s/4 and one side of the triangle would have length t/3.

The areas would then be:

\(\displaystyle \L\\A=(\frac{s}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{t}{3})^{2}\)

Now, solve [1] for whatever variable you wish and sub into the area equation. It will then be entirely in terms of one variable. Differentiate, set to 0 and solve.


For the maximum, where is the max on the interval from [0,10]?.

 

[MarkSA]

Thank you both! That was a big help. I managed to solve the parabola one on my own last night but I was still stumped on the other two.

Here is another one that is giving me trouble but that I think i've made more progress on. (we were assigned 16 of these to do)

Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3cm and 4cm if two sides of the rectangle lie along the legs.

I took the mention of the rectangle sides lying along the legs as a hint that there must be something done with that. I have a picture of a right triangle drawn and a rectangle inside it, with L being rectangle's length (which lies along the 3cm side) and W being rectangles width (which lies along the 4cm side).

I found the area of the triangle (1/2 * 3 * 4) = 6cm. If I subtract the area of the two smaller triangles formed from putting the rectangle inside from the area of the big triangle, I get:
A = 6 - [(3 - L) * W)]/2 - [(4 - W) * L]/2
Which should be the area of the rectangle.

Edit: I think I solved this finally, but it was really messy. Does this look correct? (Was there an easier way to do this one?)
Since A = lw of the rectangle, I just substituted lw for A.

LW = 6 - [(3 - L) * W)]/2 - [(4 - W) * L]/2
After a lot of simplifying I had:
W = 4 - 4/3L

After differentiating, L=1.5cm is a critical value (max). Plugging that into the Area equation I get an area of 3cm. (So that would make W=2cm)

 

[skeeter]

let one leg lie on the x-axis, from x = 0 to x = 3

let the other leg lie on the y-axis, from y = 0 to y = 4

the line between (3,0) and (0,4) is the hypotenuse ... it has the equation y = -(4/3)x + 4

the upper right corner of your rectangle lies on the line y = -(4/3)x + 4, and the area of any such rectangle is

A = xy = x[-(4/3)x + 4]

A = -(4/3)x² + 4x

dA/dx = -(8/3)x + 4

setting dA/dx = 0, you get x = 3/2, which verifies your calculator work.