Page 237 - How To Solve Word Problems In Calculus

dx 1
30 =
dt 2
dx 1
=
dt 60
1
The supply increases at the rate of thousand (approximately
60
16.67) units per week.

7. If x bears are sold at $28 each, the revenue derived from the sale is
28x.We determine the profit function by subtracting cost from
revenue.
P (x) = R(x) − C (x)

3
2
= 28x − (x − 6x + 13x + 15)
3
2
= 15x − x + 6x − 15
We proceed to find the critical value(s).
2
P (x) = 15 − 3x + 12x

2
0 = 15 − 3x + 12x
2
0 = 5 − x + 4x
2
x − 4x − 5 = 0
(x − 5)(x + 1) = 0

x − 5 = 0, x + 1 = 0

x = 5 x =−1 (x =−1 is discarded.)

The maximum daily profit is P (5) = $85.
(optional) To confirm that this is the number of teddy bears
that maximizes profit, we use the second derivative test.
P (x) =−6x + 12

P (5) =−18

Since P (5) < 0, the value x = 5 is a relative maximum.Since there

is only one relative extremum for x > 0, x = 5 corresponds to the
absolute maximum profit of $85.
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