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The following function gives the cost, in dollars, of producing
x kilograms of fertilizer:

C(x)=0.001x^(3)-0.14x^(2)+7x
What is the instantaneous rate of change of the cost when 50 kilograms are produced?
Choose 1 answer:
(A) -6.5 kilograms per dollar
(B) -6.5 dollars per kilogram
(C) 0.5 kilograms per dollar
(D) 0.5 dollars per kilogram

The following function gives the cost, in dollars, of producing $x$ kilograms of fertilizer: $\newline$ $C(x)=0.001 x^{3}-0.14 x^{2}+7 x$ $\newline$ What is the instantaneous rate of change of the cost when $50$ kilograms are produced? $\newline$ Choose $1$ answer: $\newline$ (A) $-6$ . $5$ kilograms per dollar $\newline$ (B) $-6$ . $5$ dollars per kilogram $\newline$ (C) $0$ . $5$ kilograms per dollar $\newline$ (D) $0$ . $5$ dollars per kilogram

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Solve quadratic equations: word problems

Full solution

Q. The following function gives the cost, in dollars, of producing

x

kilograms of fertilizer:

\newline

C(x)=0.001 x^{3}-0.14 x^{2}+7 x

\newline

What is the instantaneous rate of change of the cost when

50

kilograms are produced?

\newline

Choose

1

answer:

\newline

(A)

-6

.

5

kilograms per dollar

\newline

(B)

-6

.

5

dollars per kilogram

\newline

(C)

0

.

5

kilograms per dollar

\newline

(D)

0

.

5

dollars per kilogram

Calculate Derivative of $C(x)$ : To find the instantaneous rate of change, we need to calculate the derivative of the cost function $C(x)$ with respect to $x$ .
Differentiate $C(x)$ : Differentiate $C(x) = 0.001x^3 - 0.14x^2 + 7x$ to get $C'(x)$ . $\newline$ $C'(x) = 0.003x^2 - 0.28x + 7$
Find Instantaneous Rate of Change: Now, plug in $x = 50$ into the derivative to find the instantaneous rate of change at that point. $C'(50) = 0.003(50)^2 - 0.28(50) + 7$
Calculate $C'(50)$ : Calculate the value of $C'(50)$ .
$C'(50) = 0.003(2500) - 0.28(50) + 7$
$C'(50) = 7.5 - 14 + 7$
$C'(50) = 0.5$

More problems from Solve quadratic equations: word problems

Question

h

is defined over the real numbers. This table gives a few values of

h

\newline

\begin{tabular}{lllll}

\newline

x

-6

1

-6

01

-6

001

-5

9

\newline

h(x)

-0

25

-0

74

-0

98

-1

0

\newline

\newline

What is a reasonable estimate for

\lim _{x \rightarrow-6} h(x)

\newline

1

\newline

-6

\newline

-2

\newline

-1

\newline

(D) The limit doesn't exist

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