Math 1300 Written Homework #10 Solutions 4.2, Q. 24. Let y = at 2e ...

Solution: The perimeter P = 2r+2h+πr (the bottom, plus the 2 sides, plus the semicircular
arc on top). The area A = 2rh + πr 2 /2 is fixed, so we can solve that h = (A − πr 2 /2)/2r,
getting that
P = r(2 + π) + 2(A − πr 2 /2)/2r = r(2 + π/2) + A/r,
where r > 0 and r < 2A/π (so h > 0). We find critical points by differentiating P
with respect to r, yielding dP
dR = 2 + π/2 − A/r2 , which exists since r > 0, and is 0 when
r2 = A/(2 + π/2), so r = A/(2 + π/2). Since there is precisely 1 critical point, and the
derivative is negative for r < A/(2 + π/2) and positive for r > A/(2 + π/2), we have
that r = A/(2 + π/2) produces the minimum perimeter, and
h = (A − πA/(4 + π))/2 A/(2 + π/2) = √ A/(2 + π/2) 3/2 .
4.4, Q. 40. Of all rectangles with given area, A, which has the shortest diagonals?
Solution: Say the rectangle is x by y, so xy = A. Then the diagonal length we want to
mimimize is x 2 + y 2 . This is the same as minimizing s = x 2 + y 2 , which is much easier to
handle algebraically. Since y = A/x, we have
s = x 2 + A 2 /x 2 ,
where x > 0. We compute ds
dx = 2x − 2A2 /x 3 , which exists everywhere, and is zero when
x 4 = A 2 , so x = √ A. Since the derivative is negative for x < √ A and positive for x > √ A,
this is a minimum. Note that y = A/ √ A = √ A. So the square has the shortest diagonals for
a given area.
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