1. ## Maxima/minima: Still confused

Suppose that a weight is to be held 10ft below a horizontal line AB by a wire in the shape of a Y. If the points A & B are 8ft apart, a.) Estimate on your graphics calculator to the nearest foot the shortest length of wire that can be used. b.) Confirm your estimates in part a analytically... ( Ill graph it)

Suppose that a weight is to be held 10ft below a horizontal line AB by a wire in the shape of a Y.
If the points A & B are 8ft apart, find the shortest length of wire that can be used.
Code:
        A    4    C    4    B
- * - - - - + - - - - *
:   *       |     θ *
:     *     |     *
:       *   |   *
:         * | *
10           *D
:           *
:           *
:           *
-           *W

The weight is at W. .Let /CBD = θ

The length of the wire is: .L .= .AD + BD + DW

In right triangle BCD, we have: .BD = 4·secθ, CD = 4·tanθ
. . So: .DW .= .10 - 4·tanθ

And we have: .L .= .2(4·secθ) + (10 - 4·tanθ) .= .8·secθ - 4·tanθ + 10

Now you can minimize L . . .

3. See attachment for notation

x/4 = tan(theta)

4/l1 = cos(theta)

Total length of the wire is:

l = 2*l1+(10-x) = (8-4 sin(theta))/cos(theta) +10.

So if l is a minimum we have dl/dtheta=0, and:

dl/dtheta = [8 sin(theta)-4 sin^2(theta)]/cos^2(theta) -4,

so for a mininmum we have:

[8 sin(theta)-4 sin^2(theta)]/cos^2(theta) = 4

or:

8 sin(theta)= 4 (cos^2(theta)+sin^2(theta)) = 4.

Therefore:

Sin(theta) = ½, of theat=30 degrees.

Putting this back into the expression for l gives:

l = 10 + 4 sqrt(3).

RonL

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### suppose that a weight is to be held 10 ft below a horizontal line AB by a wire in the shape of a Y. if the points A and B are 8ft apart, what is the shortest total length iof wire that can be used? solve.

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