Page 123 - How To Solve Word Problems In Calculus

2
40 ± (−40) − 4(1)(200) a = 1
x =
2 b =−40
√ c = 200
40 ± 800
=
2
√
= 20 ± 10 2
√
The value x = 20 + 10 2 lies outside the interval [0, 10]. Since the
triangle’s area is 0 at the endpoints of the interval, (x = 0 and
√
x = 10), x = 20 − 10 2 must correspond to the absolute
maximum area.
The corresponding value of y is computed from
200 − 20x
y =
20 − x
√
200 − 20(20 − 10 2)
= √
20 − (20 − 10 2)
√
200 2 − 200
= √
10 2
√ √
20 2 − 20 2
= √ · √
2 2
√
40 − 20 2
=
2
√
= 20 − 10 2
Since y = x, the triangle is an isosceles right triangle. The
hypotenuse
√ √ √ √
z = x 2 = (20 − 10 2) 2 = 20 2 − 20

4. Let x be the side of the square and r the radius of the circle.
100"
4x 100 − 4x

x

r
x x

x
110