Apple introduced the first iPod in October 2001 . Sales of...

Apple introduced the first iPod in October $2001 .$ Sales of the portable music player grew slowly in the early years but began to grow rapidly after $2005 .$ But the iPod era is coming to a close. Smartphones with music and video players are replacing the iPod, along with the category of device it helped to create. Sales of the iPod worldwide from 2007 through 2011 (in millions) were approximately $$ N(t)=-2.65 t^{2}+13.13 t+39.9 \quad(0 \leq t \leq 4) $$ in year $t,$ where $t=0$ corresponds to $2007 .$ Show that the worldwide sales of the iPod peaked sometime in $2009 .$ What was the approximate largest number of iPods sold worldwide from 2007 through 2011?

Apple introduced the first iPod in October $2001 .$ Sales of the portable music player grew slowly in the early years but began to grow rapidly after $2005 .$ But the iPod era is coming to a close. Smartphones with music and video players are replacing the iPod, along with the category of device it helped to create. Sales of the iPod worldwide from 2007 through 2011 (in millions) were approximately
$$
N(t)=-2.65 t^{2}+13.13 t+39.9 \quad(0 \leq t \leq 4)
$$
in year $t,$ where $t=0$ corresponds to $2007 .$ Show that the worldwide sales of the iPod peaked sometime in $2009 .$ What was the approximate largest number of iPods sold worldwide from 2007 through 2011?

Key Concepts

Vertex of a Parabola

The vertex of a parabola is the point where the function reaches its maximum value (when the parabola opens downward) or its minimum value (when it opens upward). The vertex can be found using the formula t = -b/(2a), which identifies the input value where the objective function is optimized.

Optimization

Optimization is the process of finding the most efficient or highest/lowest possible value of a function within a given set of constraints. In the context of quadratic functions, this typically involves finding the vertex of the parabola to determine the maximum or minimum value that the function can attain, which is particularly useful in real-world applications such as maximizing sales.

Quadratic Functions

Quadratic functions are polynomial expressions of degree two that typically take the form ax² + bx + c. Their parabolic graphs can open upwards or downwards depending on the sign of the coefficient a, which leads to the existence of a maximum or minimum point, making them very useful for modeling situations with a peak or trough value.

Real-world Modeling

Real-world modeling with quadratic functions involves applying the mathematical properties of quadratic equations to describe and predict phenomena in various fields, such as economics and business. This method helps to approximate trends like growth and decline, enabling analysts to estimate peak performance or critical turning points in practical scenarios.

Transcript

00:01 So given this function, this represents the number of ipod sales from 2007 to 2011, with 2007 being time zero and 2011 being time four.

00:11 Okay.

00:12 So first let's show that the sales of ipods peaked some time.

00:18 Okay, so the sales were at the highest point.

00:21 Whenever i have a minimum or a maximum, remember that that means that my slope to my tangent line.

00:27 My slope is going to be a horizontal line.

00:31 That slope is going to be zero.

00:32 So whenever i think slope, slope at a point or tangent at a point, i'm going to think derivative.

00:39 So i can prove algebraically through my first derivative.

00:45 So first find the first derivative.

00:47 Use the power rule for each of the terms.

00:49 Bring down the exponent of 2.

00:50 This will be negative 5 .3t plus the derivative of 1313t is 1313.

00:57 And the derivative of any constant is 0.

00:59 Okay, so my first derivative, what i do with my first derivative is i'm finding when the slope is zero.

01:06 So i set my first derivative equal to zero and find that maximum point.

01:12 So what i'm going to do to get t by itself subtract 1313 from both sides, this will give me negative 1313 equals negative 5 .3t.

01:25 And when i divide by negative 5 .3 to both sides, what i'm going to get is t is approximately 2 .3 ,000, 2 .4 years...

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