Maximum Area

[footnote]

"The right triangle represents a lot with the dimensions shows (4m x 12m). Doug wants to build and fence a rectangular dog kennel inside this lot. What is the maximum possible area for the kennel?"

Let A represent the area of the dog kennel
Let l represent the length of the dog kennel
Let w represent the width of the dog kennel

Equation 1: A=lw
Equation 2:

And I'm stuck here. ^.^

 

[tkhunny]

Put your triangular lot on a set of coordinate axes. I used the short side on x and the long side on y. This gives the vertices fo the thr triangle at (0,0), (4,0), and (0,12). This also gives the equation of the hypotenuse, \(\displaystyle \frac{x}{4}+\frac{y}{12} = 1\).

Our task is to build a rectangular enclosure inside this triangle.

1) I expect one corner of the enclosure at the origin.
2) I expect the opposite corner of the enclosure to be on the line, \(\displaystyle \frac{x}{4}+\frac{y}{12} = 1\). I just don't know where.

That should get you quite a bit more down the road. Let's see what you get.

 

[soroban]

Hello, footnote!

The right triangle represents a lot with the dimensions shows (4m x 12m).
Doug wants to build and fence a rectangular dog kennel inside this lot.
What is the maximum possible area for the kennel?"

Let A represent the area of the dog kennel
Let L represent the length of the dog kennel
Let QW represent the width of the dog kennel

Equation 1: .\(\displaystyle A \,=\, LW\)

Equation 2:

And I'm stuck here.

    A *
      |   *
  4-W |       *
      |           *    E
    D * - - - - - - - *
      |       L       |   *
    W |              W|       *
      |               |           *
    B * - - - - - - - * - - - - - - - * C
              L       F     12-L

\(\displaystyle BDEF\text{ is the rectangular kennel.}\)
. . \(\displaystyle \text{Equation 1: }\:A\:=\:L\cdot W\)

\(\displaystyle BC\,=\,12,\;L \,=\, BF = DE,\;FE \,=\, 12-L\)

\(\displaystyle AB \,=\, 4,\;W \,=\, DB = EF,\;AD \,=\, 4-W\)


\(\displaystyle \text{We see that: }\;\Delta ADE \,\sim\,\Delta EFC\)

\(\displaystyle \text{Hence: }\:\frac{4-W}{L} \:=\:\frac{W}{12-L} \quad\Rightarrow\qyad L + 3W \:=\:12\)

\(\displaystyle \text{And }that\text{ is Equation 2.}\)