Page 97 - How To Solve Word Problems In Calculus

Step4
2
V (x) = 12x − 156x + 360

2
0 = 12x − 156x + 360
2
0 = 12(x − 13x + 30)
0 = 12(x − 10)(x − 3)
x = 10 x = 3

Step5
The value x = 10 may be ignored since it is outside the
15
interval 0 ≤ x ≤ .
2
x V(x)

0 0
3 486
15
0
2

3
The maximum volume of 486 in occurs when x = 3.
EXAMPLE 6
Find the dimensions of the rectangle of largest area whose
base is on the x axis and whose upper two vertices lie on the
2
parabola y = 12 − x . What is the maximum area?
Solution
Step1
Let (x, y) be the point on the parabola that meets the
corner of the rectangle.

(x, y)

y

x x

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