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Last problem of HW 13 on Applied Optimization

mbradle7

Does anyone have any advice on the last problem of HW 13 on Applied Optimization. Here’s the statement:

The frame for a kite is to be made from six pieces of wood. The four exterior pieces have been cut with the lengths indicated in the figure. To maximize the area of the kite, how long should the diagonal pieces be (in terms of a and b)?

mark

First off, let’s label the unknowns:

Note that our objective function (namely, the area we are trying to maximize) can be expressed as a function of three variables:

A(x,y,z) = x\times z + y\times z.

Also, there are equations relating x, y, and z:

x^2 + z^2 = a^2 \text{ and } y^2 + z^2 = b^2

so that

x = \sqrt{a^2 - z^2} \text{ and } y = \sqrt{b^2-z^2}.

Thus, our area function can be expressed as a function of the single variable z:

A(z) = z\sqrt{a^2-z^2} + z\sqrt{b^2-z^2}.

So that’s the function we’ve got to maximize subject to 0 < z < a < b.

It doesn’t start out so bad:

\begin{aligned} A'(z) &= \sqrt{a^2-z^2}-\frac{z^2}{\sqrt{a^2-z^2}} + \sqrt{b^2-z^2}-\frac{z^2}{\sqrt{b^2-z^2}} \\ &= \frac{a^2-2 z^2}{\sqrt{a^2-z^2}} - \frac{b^2-2 z^2}{\sqrt{b^2-z^2}} \stackrel{?}{=} 0, \end{aligned}

which we can write

\frac{a^2-2 z^2}{\sqrt{a^2-z^2}} = \frac{b^2-2 z^2}{\sqrt{b^2-z^2}}.

Now, we square both sides to get rid of the square roots

\frac{(a^2-2 z^2)^2}{a^2-z^2} = \frac{(b^2-2 z^2)^2}{b^2-z^2},

which we can write as

(a^2-2 z^2)^2(b^2-z^2) - (b^2-2 z^2)^2(a^2-z^2) = 0.

Fortunately, there is a lot of cancellation when we expand the left and, in fact, we get

z^2b^4-z^2a^4 + a^4 b^2 - a^2 b^4 = 0.

Now, that’s pretty easy to solve for z and, in fact, we get

z = \sqrt{\left(a^2b^4 - a^4b^2\right)/\left(b^4 - a^4\right)}.

Finally, that tells us the value of z that we need and, looking back at the picture, we can figure out what the lengths of the diagonals are in terms of z.