2. Ex 4 p. 221 Finding minimum length. Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire? x 30-x y z 28’ 12’ W = y + z Primary, W is wire x 2 +12 2 = y 2 Secondary (30 – x) 2 + 28 2 = z 2 Secondary Solve for y, solve for z, replace so W is an equation in terms of x.
3. W = y + z, x 2 +12 2 = y 2 (30 – x) 2 + 28 2 = z 2 Let W’ = 0 to get critical #’s Square both sides Stake at 9 feet from short pole
4. You can confirm your math by graphing the Primary equation and find minimum of W. In each of the previous examples, the extreme value occurred at a critical number. This often happens, but remember that an extreme value can occur at an endpoint of an interval as well, as shown in the next example. Ex 5 p. 222 An endpoint maximum Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much for the wire to enclose the maximum area?
5. x x Area = x 2 Perimeter = 4x r Area = π r 2 Circum = 2 π r Total Area = x 2 + π r 2 Primary Wire is 4 ft, so 4 = 4x + 2 π r Secondary Solving for r, Feasible domain is 0 ≤ x ≤ 1, controlled by perimeter of square
6. 0 ≤ x ≤ 1 If we evaluate A(0) ≈ 1.273, A(.56) ≈ .56, and A(1) = 1, we can see that the greatest area occurs when x = 0, or in other words, all the wire is used to form the circle, none is used to form a square!
