https://r2.erweima.ai/mathbot/image/17265497307399fxupl2i.pdf

Let's break down and solve the math exercises from the provided file. Here's a step-by-step analysis:

### **Exercice 1:**
A box without a lid is constructed from a rectangular metal sheet with dimensions 18 cm by 24 cm. The problem involves cutting squares of side \( x \) cm from each corner and folding up the edges to create the box. The volume of the box \( V(x) \) is defined for \( x \in [0,9] \).

#### **1) Justify the volume formula:**
We are asked to justify that:
\[
V(x) = 4x^3 - 84x^2 + 432x
\]
- The box's base dimensions, after cutting squares of side \( x \), will be:
  - Length: \( 24 - 2x \)
  - Width: \( 18 - 2x \)
- The height of the box is \( x \).
Thus, the volume \( V(x) \) is given by:
\[
V(x) = (24 - 2x)(18 - 2x)x
\]
Expanding this expression:
\[
V(x) = (432 - 48x - 36x + 4x^2)x
= 432x - 84x^2 + 4x^3
\]
Hence, \( V(x) = 4x^3 - 84x^2 + 432x \), as required.

#### **2) Derivative of \( V(x) \):**
To find the derivative \( V'(x) \), we differentiate:
\[
V'(x) = \frac{d}{dx}(4x^3 - 84x^2 + 432x)
= 12x^2 - 168x + 432
\]
Thus, \( V'(x) = 12x^2 - 168x + 432 \).

#### **3) Variation table for \( V(x) \):**
We need to find critical points by solving \( V'(x) = 0 \):
\[
12x^2 - 168x + 432 = 0
\]
Dividing by 12:
\[
x^2 - 14x + 36 = 0
\]
The discriminant is:
\[
\Delta = (-14)^2 - 4(1)(36) = 196 - 144 = 52
\]
Thus, the roots are:
\[
x = \frac{14 \pm \sqrt{52}}{2} = \frac{14 \pm 2\sqrt{13}}{2}
= 7 \pm \sqrt{13}
\]
Approximate roots:
\[
x_1 \approx 7 + 3.6 = 10.6 \quad (\text{not valid, as } x \leq 9)
\]
\[
x_2 \approx 7 - 3.6 = 3.4
\]
Now, we can create a variation table based on the behavior of \( V'(x) \), indicating increasing and decreasing intervals.

#### **4) Maximum volume:**
The maximum volume occurs at the critical point \( x \approx 3.4 \). We would substitute this value back into \( V(x) \) to find the exact volume.

#### **5) Volume constraint:**
To determine if the box can have a volume \( \geq 650 \text{ cm}^3 \), we solve \( V(x) \geq 650 \) using the volume formula and analyze the inequality.

### **Exercice 2:**
This involves analyzing two functions, \( f(x) \) and \( g(x) \).

#### **1a) Derivative of \( f(x) \):**
Given \( f(x) = x^3 + 3x^2 - 9x - 1 \), we differentiate:
\[
f'(x) = 3x^2 + 6x - 9
\]

#### **1b) Sign of \( f'(x) \):**
Solve \( f'(x) = 0 \):
\[
3x^2 + 6x - 9 = 0 \quad \Rightarrow \quad x^2 + 2x - 3 = 0
\]
The roots are:
\[
x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} = \frac{-2 \pm \sqrt{16}}{2}
= \frac{-2 \pm 4}{2}
\]
Thus, \( x = 1 \) or \( x = -3 \). Use this to find the intervals where \( f'(x) \) is positive or negative.

#### **1c) Tangent at \( x = -1 \):**
Find the equation of the tangent line at \( x = -1 \). The slope is \( f'(-1) \) and the point on the curve is \( f(-1) \).

#### **2a & 2b) Function \( g(x) \):**
Analyze the quadratic function \( g(x) = 10x^2 + 8x + 8 \) and its discriminant to confirm that \( f \) and \( g \) have the same tangent at \( x = -1 \).

---

Let me know if you need details on any specific part or solution verification!

### Follow-up questions:
1. What is the volume of the box when \( x = 3.4 \)?
2. Can you solve \( V(x) \geq 650 \)?
3. How do you create a variation table for a function?
4. What is the slope of the tangent line for \( f(x) \) at \( x = -1 \)?
5. How does the discriminant affect the shape of a quadratic function?

### Tip:
When solving optimization problems, always check the critical points and boundaries to ensure you've found the maximum or minimum values.

An industrialist wants to manufacture a box without a lid from a metal sheet measuring 18 cm by 24 cm. The process involves cutting squares of side x cm from each corner and folding the sides vertically. The volume of the box is given as a function V(x) over the interval [0, 9]. Tasks include justifying the volume formula, finding the derivative, creating a variation table, determining the value of x that maximizes the volume, and checking if the box can have a volume greater than or equal to 650 cm³.

Solve a volume optimization problem involving cutting squares from a metal sheet to form a box. This exercise covers polynomial functions, derivatives, and finding maxima. Learn to justify volume formulas and create a variation table for optimization.

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