Exercise 2 (5 points)

Part A

A craftsman creates chocolate candies whose shape recalls the profile of the local mountain. The base of such a candy is modeled by the shaded surface defined in an orthonormal coordinate system with unit 1 cm.
This surface is bounded by the x-axis and the graph denoted $\mathscr { C } _ { f }$ of the function $f$ defined on $[ - 1 ; 1 ]$ by: $$f ( x ) = \left( 1 - x ^ { 2 } \right) \mathrm { e } ^ { x } .$$
The objective of this part is to calculate the volume of chocolate needed to manufacture a chocolate candy.

1. a. Justify that for all $x$ belonging to the interval $[ - 1 ; 1 ]$ we have $f ( x ) \geqslant 0$. b. Show using integration by parts that: $$\int _ { - 1 } ^ { 1 } f ( x ) \mathrm { d } x = 2 \int _ { - 1 } ^ { 1 } x \mathrm { e } ^ { x } \mathrm {~d} x .$$
2. The volume $V$ of chocolate, in $\mathrm { cm } ^ { 3 }$, needed to manufacture a candy is given by: $$V = 3 \times S$$ where $S$ is the area, in $\mathrm { cm } ^ { 2 }$, of the colored surface. Deduce that this volume, rounded to $0.1 \mathrm {~cm} ^ { 3 }$, equals $4.4 \mathrm {~cm} ^ { 3 }$.

Part B

We now consider the profit realized by the craftsman on the sale of these chocolate candies as a function of the weekly sales volume.
This profit can be modeled by the function $B$ defined on the interval $[ 0.01 ; + \infty [$ by: $$B ( q ) = 8 q ^ { 2 } [ 2 - 3 \ln ( q ) ] - 3 .$$
The profit is expressed in tens of euros and the quantity $q$ in hundreds of candies. We admit that the function $B$ is differentiable on $\left[ 0.01 ; + \infty \left[ \right. \right.$. We denote $B ^ { \prime }$ its derivative function.

1. a. Determine $\lim _ { q \rightarrow + \infty } B ( q )$. b. Show that, for all $q \geqslant 0.01 , B ^ { \prime } ( q ) = 8 q ( 1 - 6 \ln ( q ) )$. c. Study the sign of $B ^ { \prime } ( q )$, and deduce the direction of variation of $B$ on $[ 0.01 ; + \infty [$. Draw the complete variation table of function $B$. d. What is the maximum profit, to the nearest euro, that the craftsman can expect?
2. a. Show that the equation $B ( q ) = 10$ has a unique solution $\beta$ on the interval $[1.2; + \infty [$. Give an approximate value of $\beta$ to $10 ^ { - 3 }$ near. b. We admit that the equation $B ( q ) = 10$ has a unique solution $\alpha$ on $[ 0.01 ; 1.2 [$. We are given $\alpha \approx 0.757$. Deduce the minimum and maximum number of chocolate candies to sell to achieve a profit greater than 100 euros.

For a positive number $a$, the function $f ( x ) = \int _ { 0 } ^ { x } ( a - t ) e ^ { t } \, d t$ has a maximum value of 32. Find the area enclosed by the curve $y = 3 e ^ { x }$ and the two lines $x = a$ and $y = 3$. [4 points]

We propose to calculate the area $\mathscr{A}$ of the domain $\mathscr{H}$ of $\mathbb{R}^2$ containing all the points $w(n, t)$ when $n$ ranges over $\mathbb{N}^*$ and $t$ ranges over $I = \left[\frac{\pi}{4}, \frac{7\pi}{4}\right]$. This domain is bounded by two parametrized arcs defined by $$z : I \rightarrow \mathbb{C}, t \mapsto \rho(t) \mathrm{e}^{\mathrm{i}\theta(t)} = \sqrt{1 + 3\sin^2 t}\, \mathrm{e}^{\mathrm{i}\left(\arctan(2\tan t) + \pi \mathrm{E}\left(\frac{t}{\pi} + \frac{1}{2}\right)\right)}$$ $$v : I \rightarrow \mathbb{C}, t \mapsto \sqrt{1 + 3\sin^2 t}\left(1 + \frac{1}{4}\sin\left(\frac{2}{3}\left(t - \frac{\pi}{4}\right)\right)\right) \mathrm{e}^{\mathrm{i}\left(\arctan(2\tan t) + \pi \mathrm{E}\left(\frac{t}{\pi} + \frac{1}{2}\right)\right)}$$
III.D.1) Recall the statement of the Green-Riemann theorem. Explain how this theorem translates in the case of an area calculation. III.D.2) Recall the formula giving the scalar product of two complex numbers. Deduce the expression of the scalar product $\langle u \circ v(t), v'(t) \rangle$, when $u$ and $v$ are the applications $u : \mathbb{C} \rightarrow \mathbb{C}, z \mapsto \mathrm{i}z$ and $v : t \mapsto \sigma(t) \mathrm{e}^{\mathrm{i}\mu(t)}$, where $\sigma$ and $\mu$ are two functions defined on an interval $J$ of $\mathbb{R}$, with real values and of class $C^1$. III.D.3) If $d(t) = \arctan(2\tan(t))$, simplify $\frac{1}{2}\left(1 + 3\sin^2 t\right) d'(t)$. III.D.4) Deduce from the previous questions an expression of $\mathscr{A}$ in the form of an integral. Simplify this integral using the identity obtained in III.D.3). Finally, calculate $\mathscr{A}$.

Let $a$ be a positive real number. Let P denote the point of intersection of the following two curves
$$\begin{aligned} & C _ { 1 } : y = \frac { 3 } { x } \\ & C _ { 2 } : y = \frac { a } { x ^ { 2 } } , \end{aligned}$$
and let $\ell$ denote the tangent to $C _ { 2 }$ at P. Then we are to find the area $S$ of the region bounded by $C _ { 1 }$ and $\ell$.
Since the coordinates of P are $\left( \frac { a } { \mathbf { N } } , \frac { \mathbf { O } } { a } \right)$, the equation of $\ell$ is
$$y = - \frac { \mathbf { P Q } } { a ^ { 2 } } x + \frac { \mathbf { R S } } { a }$$
When we set
$$p = \frac { a } { \mathbf { T } } , \quad q = \frac { a } { \mathbf { U } } \quad ( p < q )$$
$S$ is obtained by calculating
$$S = [ \mathbf { V } ] _ { p } ^ { q }$$
where $\mathbf{V}$ is the appropriate expression from among (0) $\sim$ (5) below.
(0) $\frac { 18 } { a ^ { 2 } } x ^ { 2 } - \frac { 27 } { a } x + 3 \log | x |$
(1) $\frac { 9 } { a ^ { 2 } } x ^ { 2 } - \frac { 9 } { a } x + 3 \log | x |$
(2) $- \frac { 27 } { a ^ { 2 } } x ^ { 2 } + \frac { 18 } { a } x - 3 \log | x |$
(3) $- \frac { 27 } { a ^ { 2 } } x ^ { 2 } + \frac { 27 } { a } x - 3 \log | x |$
(4) $\frac { 27 } { a ^ { 2 } } x ^ { 2 } - \frac { 27 } { a } x + 3 \log | x |$
(5) $- \frac { 18 } { a ^ { 2 } } x ^ { 2 } + \frac { 27 } { a } x - 3 \log | x |$
Hence we obtain
$$S = \frac { \mathbf { W } } { \mathbf { X } } - 3 \log \mathbf { Y }$$

Let $a > 0$. Consider the region of a plane bounded by the curve $y = \sqrt { x } e ^ { - x }$, the $x$-axis, and the straight line $x = a$ which passes through the point $\mathrm{ A }( a , 0 )$, and let $V$ be the volume of the solid obtained by rotating this region once about the $x$-axis.
(1) $V$ is expressed as a function in $a$ by
$$V = - \frac { \pi } { 4 } \left\{ ( \mathbf { N } a + \mathbf { O } ) e ^ { - \mathbf { P } a } - \mathbf { Q R } \right\} .$$
(2) Suppose that the point A starts at the origin and moves along the $x$-axis in the positive direction and that its speed at $t$ seconds is $4t$. Then the rate of change of $V$ at $t$ seconds is
$$\frac { d V } { d t } = \mathbf { R } \pi t ^ { \mathbf { S } } e ^ { - \mathbf { T } t ^ { \mathbf { U } } } .$$
This rate of change is maximized at
$$t = \frac { \sqrt { \mathbf { V } } } { 4 } ,$$
and the value of $V$ at this time is
$$V = - \frac { \pi } { 8 } \left( \mathbf { W } e ^ { - \frac { \mathbf { X } } { \mathbf { Y } } } - \mathbf { Z } \right) .$$

Consider the two functions
$$y = x \log a x , \tag{1}$$ $$y = 2 x - 3 , \tag{2}$$
where $a > 0$, and where $\log$ is the natural logarithm.
(1) Let us find $a$ such that the graph of (1) is tangent to the graph of (2).
The equation of the tangent to the graph of (1) at the point $( t , t \log a t )$ is $\mathbf { A }$ (for A, choose the correct answer from among choices (0) $\sim$ (3) below). (0) $y = ( \log a t + 1 ) x - t$
(1) $y = ( \log a t + a ) x - t$
(2) $y = ( a \log t + 1 ) x + t$
(3) $y = ( a \log t + a ) x + t$
Hence, the graph of (1) is tangent to the graph of (2) when $a = \frac { e } { \square \mathbf{B} }$, and the coordinates of the tangent point are ( $\mathbf { C }$, $\mathbf { D }$ ).
(2) When $a = \frac { e } { \mathbf{B} }$, function (1) is minimized at $x = \square e ^ { - \mathbf { F } }$, and in this case the minimum value is $- \mathbf { G } \cdot e ^ { - \mathbf { H } }$.
(3) When $a = \frac { e } { \mathbf{B} }$, let us find the area $S$ of the region bounded by the graphs of (1) and (2) and the $x$-axis.
For the following indefinite integral, we have
$$\int x \log a x \, d x = \square + C , \quad \text { where } C \text { is an integral constant }$$
(for I, choose the correct answer from among (0) $\sim$ (3) below). (0) $\frac { 1 } { 2 } x ^ { 2 } \log a x - \frac { 1 } { 2 } x ^ { 2 }$
(1) $2 x ^ { 2 } \log a x - 2 x ^ { 2 }$
(2) $\frac { 1 } { 2 } x ^ { 2 } \log a x - \frac { 1 } { 4 } x ^ { 2 }$
(3) $2 x ^ { 2 } \log a x - 4 x ^ { 2 }$
Hence we obtain
$$S = \frac { \mathbf { J } } { \mathbf { K } }$$