/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 The sales of functional food pro... [FREE SOLUTION] | 91Ó°ÊÓ

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The sales of functional food products-those that promise benefits beyond basic nutrition-have risen sharply in recent years. The sales (in billions of dollars) of foods and beverages with herbal and other additives is approximated by the function \(S(t)=0.46 t^{3}-2.22 t^{2}+6.21 t+17.25 \quad(0 \leq t \leq 4)\) where \(t\) is measured in years, with \(t=0\) corresponding to the beginning of 1997 . Show that \(S\) is increasing on the interval \([0,4]\). Hint: Use the quadratic formula.

Short Answer

Expert verified
To show that the function \(S(t)\) is increasing on the interval \([0,4]\), we first find the first derivative of the function: \(S'(t) = 1.38t^2 - 4.44t + 6.21\). To analyze the sign of the first derivative, we use the quadratic formula and find out that the discriminant is negative, which means that there are no real roots. Since the coefficient of \(t^2\) is positive, the first derivative is always positive for any \(t\in\mathbb{R}\). Thus, we conclude that \(S(t)\) is increasing on the interval \([0, 4]\).

Step by step solution

01

Find the first derivative of S(t)

To find the first derivative of the function \(S(t)\), we'll use basic power rule differentiation. \[ S(t) = 0.46 t^3 - 2.22 t^2 + 6.21 t + 17.25 \] \[ S'(t) = \frac{d}{dt}(0.46 t^3 - 2.22 t^2 + 6.21 t + 17.25) \] Apply power rule for differentiation: \[ S'(t) = 0.46(3) t^{3-1} - 2.22(2) t^{2-1} + 6.21(1) t^{1-1} \] \[ S'(t) = 1.38t^2 - 4.44t + 6.21 \]
02

Analyze the sign of the first derivative using the quadratic formula

We want to show that \(S(t)\) is always positive for \( 0\leq t \leq 4\). Now we have the first derivate as a quadratic expression. To analyze its sign, we can look for the real roots (if any) using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Apply the quadratic formula with \(a = 1.38\), \(b = -4.44\), and \(c = 6.21\): \[ t = \frac{4.44 \pm \sqrt{(-4.44)^2 - 4(1.38)(6.21)}}{2(1.38)} \] Calculate the value inside the square root: \[ Discriminant = b^2-4ac = (-4.44)^2 - 4(1.38)(6.21) = 19.7136 - 34.3032 = -14.5896 \] The discriminant is negative, which means that the quadratic function has no real roots. Also, since \(a=1.38 > 0\), we know that the parabolic function is opening upwards. In this case, we can conclude that the first derivative \(S'(t)\) is always positive for any \(t \in \mathbb{R}\).
03

Conclude that S(t) is increasing on the interval [0,4]

Since we found that the first derivative of the function \(S'(t) = 1.38t^2 - 4.44t + 6.21\) is always positive for any \(t \in \mathbb{R}\), we can conclude that \(S(t)\) is an increasing function on the interval \([0, 4]\).

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Key Concepts

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

Derivative Analysis
In calculus, derivative analysis helps us understand the behavior of a function, particularly its rate of change. When dealing with functions like sales approximations, it's important to know if they are increasing or decreasing, which is where the first derivative comes into play.
To find the first derivative of a function, such as the sales function, we apply differentiation rules like the power rule, which involves multiplying the exponent by the coefficient and then reducing the exponent by one. This results in a new function that shows how the original function changes at any given point.
For the sales function \( S(t) = 0.46 t^3 - 2.22 t^2 + 6.21 t + 17.25 \), the derivative is \( S'(t) = 1.38t^2 - 4.44t + 6.21 \). Here, \( S'(t) \) helps determine if the sales are rising or falling over time. Analyzing the sign of this derivative tells us whether the sales function is increasing or decreasing on the specified interval. If the entire first derivative is positive, as it is from the calculation, then the original function is increasing.
This simple yet profound analysis reveals the growth trend of sales over the specific timeline and is crucial for businesses planning future strategies.
Quadratic Formula
The quadratic formula is a versatile and essential tool for finding the roots of quadratic equations, which are polynomials of degree two. A quadratic function in standard form is given by \( ax^2 + bx + c = 0 \). The roots or solutions of this equation are found through the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This formula is applicable when the expression inside the square root, called the discriminant \( b^2 - 4ac \), is evaluated. If the discriminant is positive, the quadratic equation has two distinct real roots. If it is zero, there is exactly one real root. However, if the discriminant is negative, as in our original problem, it indicates that no real roots exist and the parabola does not intersect the x-axis.
This result is particularly important because it helps us determine a function's range or verify the sign of the expression over a desired interval. In the context of our sales function's derivative \( S'(t) = 1.38t^2 - 4.44t + 6.21 \), a negative discriminant confirms that the derivative remains positive over its domain, implying a consistent increase in sales over time.
Increasing Functions
Increasing functions are functions whose output values rise as the input values increase. This property indicates a positive trend and is fundamental in fields like economics to understand growth metrics.
For a function to be increasing on an interval, its derivative should be positive across the entire interval. Derivative analysis, as seen earlier, helps confirm this. When the first derivative \( S'(t) \) of a function is consistently positive, as with the sales function \( S'(t) = 1.38t^2 - 4.44t + 6.21 \), it concludes that the function is strictly increasing over the given time period from \( t = 0 \) to \( t = 4 \).
Understanding whether a function is increasing is crucial for businesses and analysts. It helps in decision-making by providing insight into trends, enabling predictions, and allowing stakeholders to see the bigger picture. By knowing that the function is increasing, companies can confidently infer that sales will likely grow, which can influence investment and marketing strategies.

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Most popular questions from this chapter

A stone is thrown straight up from the roof of an \(80-\mathrm{ft}\) building. The height (in feet) of the stone at any time \(t\) (in seconds), measured from the ground, is given by $$ h(t)=-16 t^{2}+64 t+80 $$ What is the maximum height the stone reaches?

It has been estimated that the total production of oil from a certain oil well is given by $$ T(t)=-1000(t+10) e^{-0.1 t}+10,000 $$ thousand barrels \(t\) yr after production has begun. Determine the year when the oil well will be producing at maximum capacity.

The average annual price of single-family homes in Massachusetts between 1990 and 2002 is approximated by the function \(P(t)=-0.183 t^{3}+4.65 t^{2}-17.3 t+200 \quad(0 \leq t \leq 12)\) where \(P(t)\) is measured in thousands of dollars and \(t\) is measured in years, with \(t=0\) corresponding to 1990 . In what year was the average annual price of single-family homes in Massachusetts lowest? What was the approximate lowest average annual price? Hint: Use the quadratic formula.

The number of major crimes committed in the city of Bronxville between 2000 and 2007 is approximated by the function $$ N(t)=-0.1 t^{3}+1.5 t^{2}+100 \quad(0 \leq t \leq 7) $$ where \(N(t)\) denotes the number of crimes committed in year \(t(t=0\) corresponds to 2000 ). Enraged by the dramatic increase in the crime rate, the citizens of Bronxville, with the help of the local police, organized "Neighborhood Crime Watch" groups in early 2004 to combat this menace. Show that the growth in the crime rate was maximal in 2005 , giving credence to the claim that the Neighborhood Crime Watch program was working.

Find the dimensions of a rectangle of area 144 sq \(\mathrm{ft}\) that has the smallest possible perimeter.

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