Page 128 - How To Solve Word Problems In Calculus

√ By the theorem of Pythagoras,
3 2
2
x
2
2
A = xy + x h + = x ; it follows that
4 2 2
Since the perimeter of the window h = x − x = 3 x and h =
2
2
2
is 600 cm, √ 4 4
3
3x + 2y = 600 2 x. The area of the equilat-
√
2y = 600 − 3x eral triangle = 1 3 x . 2
3 2 xh = 4
y = 300 − x
2
The minimum value of x is 0 and the maximum value, which occurs
when y = 0, is 200. Hence 0 ≤ x ≤ 200.
√

3 3
A(x) = x 300 − x + x 2
2 4
√
3 3
2
= 300x − x + x 2
2 4
Differentiating and setting the derivative equal to 0, we obtain
√
3

A (x) = 300 − 3x + x
2
√
3
0 = 300 − 3x + x
2
√
3
3x − x = 300
2
√
6x − x 3 = 600
√
(6 − 3)x = 600
600
x = √
6 − 3
2 √ 2
600 600 3 600 3 600
A √ = 300 √ − √ + √
6 − 3 6 − 3 2 6 − 3 4 6 − 3
90,000
= √ (verify)
6 − 3
≈ 21,087 cm 2
√
3 600

(optional) Since A (x) =−3 + < 0, x = √ yields a
2 6 − 3
115