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Daily oil production by Pemex, Mexico's national oil company, can be approximated by \(q(t)=-0.022 t^{2}+0.2 t+2.9\) million barrels \(\quad(1 \leq t \leq 9)\) where \(t\) is time in years since the start of \(2000 .^{54}\) At the start of 2008 the price of oil was \(\$ 90\) per barrel and increasing at a rate of \(\$ 80\) per year. \(^{55}\) How fast was Pemex's oil (daily) revenue changing at that time?

Short Answer

Expert verified
At the start of 2008, Pemex's daily oil revenue was decreasing at a rate of $70 million per year.

Step by step solution

01

Find the derivative of the production function dq/dt

The production function is given as \(q(t) = -0.022t^2 + 0.2t + 2.9\). We need to find its derivative with respect to time t, i.e., \(dq/dt\). Using power rule of differentiation, we have: \( \frac{dq}{dt} = \frac{d}{dt}(-0.022t^2 + 0.2t + 2.9) \) \( \frac{dq}{dt} = -0.044t + 0.2 \)
02

Set price function P(t)

The price of oil at the start of 2008 was given as \(90 per barrel, and it was increasing at a rate of \)80 per year. We can denote the price function as \(P(t)\), where \(t\) is the time in years since the start of 2000. Since the price is increasing linearly with time, we can represent it as: \(P(t) = 90 + 80(t - 8)\) Here, at \(t=8\), \(P(t)\) becomes 90 which is the known price at that moment.
03

Find the derivative of the price function dP/dt

To find the rate of change of the price of oil with respect to time, we need to find the derivative of the price function, dP/dt. \( \frac{dP}{dt} = \frac{d}{dt}(90 + 80(t - 8)) \) \( \frac{dP}{dt} = 80 \)
04

Find the derivative of the revenue function dR/dt

To find the rate of change of the revenue, we need to find the derivative of the revenue function, dR/dt. The revenue function is given by the product of price function and production function, i.e. \(R(t) = P(t) * q(t)\). We will use the product rule and the chain rule here, which states that: \(\frac{dR}{dt} = P(t) \cdot \frac{dq}{dt} + q(t) \cdot \frac{dP}{dt}\) Now, substituting the time t=8 (start of 2008): \( \frac{dR}{dt} = P(8) \cdot \frac{dq}{dt}(8) + q(8) \cdot \frac{dP}{dt}(8) \) \( \frac{dR}{dt} = (90) \cdot (-0.044(8) + 0.2) + (-0.022(8)^2 + 0.2(8) + 2.9) \cdot (80) \)
05

Calculate the value of dR/dt

Finally, we will compute the value of dR/dt: \(\frac{dR}{dt} = 90\cdot(-0.352+0.2)+(-0.022\cdot(8)^2+0.2\cdot 8+2.9)\cdot 80\) \(\frac{dR}{dt} = 90(-0.152)+(-0.704)\cdot 80\) \(\frac{dR}{dt} = -13.68 - 56.32\) \(\frac{dR}{dt} = -70\) At the start of 2008, Pemex's daily oil revenue was decreasing at a rate of $70 million per year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
In calculus, a derivative is a fundamental concept that represents the rate at which a quantity changes. Specifically, it measures how a function's output value changes as its input changes. For a function \(f(x)\), the derivative is denoted as \(f'(x)\) or \(\frac{df}{dx}\). It provides critical information about the function's behavior, such as identifying where it increases or decreases.

To determine a derivative, differentiation techniques are used, which involve calculating the limit of the ratio of change in the function over a change in its input as the change approaches zero. In practice, derivatives are crucial in various fields, including physics, engineering, and economics, as they offer insights into trends and changes over time.
Revenue function
The revenue function is an essential concept in business and economics, representing how revenue changes relative to varying inputs, such as quantity or sales price. Mathematically, if \(q(t)\) represents the quantity of goods sold and \(P(t)\) stands for the price per unit, then the revenue function \(R(t)\) is expressed as \(R(t) = q(t) \cdot P(t)\).

This function is pivotal in determining how much money a company makes in relation to different factors, such as time or market conditions. It helps businesses forecast earnings and understand the impact of different variables on their overall revenue. Analyzing the revenue function allows businesses to make data-driven decisions that maximize profitability.
Product rule
The product rule is a crucial technique in differentiation used when differentiating products of two functions. If you have a function \(R(t) = f(t) \cdot g(t)\), the product rule states that the derivative \(\frac{dR}{dt}\) is given by:

\[\frac{dR}{dt} = f(t) \frac{dg}{dt} + g(t) \frac{df}{dt}\]

This rule is particularly helpful when dealing with complex revenue functions where both production level and price might change over time. Applying the product rule effectively allows you to break down the problem into simpler parts, calculate the derivatives individually, and then appropriately combine them to find the overall rate of change of the product of two functions.
Differentiation
Differentiation is a process in calculus used to find the derivative of a function. It is the operation of calculating the rate of change of a function's output value with respect to a change in its input value. This technique involves principles such as the power rule, product rule, and chain rule, each suited for different types of functions.

For example, the power rule facilitates finding the derivative of power functions, while the product rule assists with the differentiation of products of functions. Differentiation is significant in many real-world applications, such as predicting market trends and optimizing complex systems. It provides tools to analyze how changes in one quantity affect another, thus offering strategic insights into dynamic processes.

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Most popular questions from this chapter

You have been hired as a marketing consultant to Big Book Publishing, Inc., and you have been approached to determine the best selling price for the hit calculus text by Whiner and Istanbul entitled Fun with Derivatives. You decide to make life easy and assume that the demand equation for Fun with Derivatives has the linear form \(q=m p+b\), where \(p\) is the price per book, \(q\) is the demand in annual sales, and \(m\) and \(b\) are certain constants you'll have to figure out. a. Your market studies reveal the following sales figures: when the price is set at \(\$ 50.00\) per book, the sales amount to 10,000 per year; when the price is set at \(\$ 80.00\) per book, the sales drop to 1000 per year. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes annual revenue and predict what Big Book Publishing, Inc.'s annual revenue will be at that price.

A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 10 centimeters. Assuming the balloon is filled with helium at a rate of 10 cubic centimeters per second, calculate how fast the radius is growing at the instant it pops. (The volume of a sphere of radius \(r\) is \(V=\frac{4}{3} \pi r^{3}\).) HINT [See Example 1.]

The number \(N\) of employees and the total floor space \(S\) of your company are both changing with time. Show that the percentage rate of change of square footage per employee equals the percentage rate of change of \(S\) minus the percentage rate of change of \(N\). (The percentage rate of change of a quantity \(Q\) is \(\left.Q^{\prime}(t) / Q(t) .\right)\)

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If we regard position, \(s\), as a function of time, \(t\), what is the significance of the third derivative, \(s^{\prime \prime \prime}(t) ?\) Describe an everyday scenario in which this arises.

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