Q.A square piece of tin of side $18$ cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

The problem gives a cone with radius $\sqrt{3}$ and semivertical angle $\frac{\pi}{3}$. The semivertical angle is the angle between the axis and the slant edge. That means in a vertical cross-section through the axis, the cone looks like an isosceles triangle with base $2\sqrt{3}$ and height $h$ such that $\tan(\pi/3) = \frac{\text{radius}}{\text{height}} = \frac{\sqrt{3}}{h}$. Since $\tan(\pi/3) = \sqrt{3}$, we get $\sqrt{3} = \frac{\sqrt{3}}{h}$, so $h = 1$. The cone’s height is $1$.

Now imagine a right circular cylinder inscribed in this cone — its axis coincides with the cone’s axis, and its top face touches the slant surface. In the cross-section, the cylinder appears as a rectangle inside the triangle. Let the cylinder’s radius be $r$ and its height be $H$. The key is that the top corners of the rectangle lie on the slant edges of the triangle.

1. Relate $r$ and $H$ using similar triangles. In the cross-section, consider the smaller triangle above the cylinder: its base is $r$ (half the cylinder’s diameter) and its height is $1 - H$ (the distance from the cylinder’s top to the cone’s apex). This small triangle is similar to the whole triangle (base $\sqrt{3}$, height $1$). So:

Hence $r = \sqrt{3}\,(1 - H)$.

2. Write the volume of the cylinder.

   Volume $V = \pi r^2 H = \pi \left[\sqrt{3}(1-H)\right]^2 H = 3\pi (1-H)^2 H$.

3. Maximise $V$ with respect to $H$. …

We are asked for the absolute maximum — the highest value the function reaches anywhere on the given closed interval. A cubic polynomial is continuous everywhere, so on a closed interval it must attain both a maximum and a minimum. The candidates are the endpoints and any points where the derivative is zero (critical points) inside the interval.

The derivative is $f'(x) = 6x^2 - 6x - 36$. Factor it: $6(x^2 - x - 6) = 6(x-3)(x+2)$. So $f'(x)=0$ at $x=3$ and $x=-2$. Both lie in $[-3,3]$, so they are valid critical points.

Now evaluate $f$ at all candidates:

1. At $x=-3$ (left endpoint):

   $f(-3) = 2(-27) - 3(9) - 36(-3) + 9 = -54 - 27 + 108 + 9 = 36$.

2. At $x=-2$ (critical point):

   $f(-2) = 2(-8) - 3(4) - 36(-2) + 9 = -16 - 12 + 72 + 9 = 53$.

3. At $x=3$ (critical point and right endpoint):

   $f(3) = 2(27) - 3(9) - 36(3) + 9 = 54 - 27 - 108 + 9 = -72$.

The values are $36$, $53$, and $-72$. The largest is $53$ — but wait, that's not among the options. Let's check again.

$f(-2) = 2(-8) - 3(4) - 36(-2) + 9 = -16 - 12 + 72 + 9$. …

Left branch $(x\le1):\ f(x)=1+6x-3x^2$. Its derivative $f'(x)=6-6x=0$ at $x=1$, and the parabola opens downward, so on $(-\infty,1]$ the maximum is at $x=1$:

$f(1)=1+6-3=4.$ …

A local maximum of a function $f(x)$ at a point $(x_0, y_0)$ provides two critical pieces of information:

1. The point lies on the curve: The function's value at $x_0$ is $y_0$. This means $f(x_0) = y_0$.
2. The slope of the tangent is zero: At a local maximum (or minimum), the tangent line to the curve is horizontal. The slope of this tangent is given by the first derivative, so $f'(x_0) = 0$.

We will use these two conditions to set up a system of equations for $a$ and $b$, and then solve for them.

1. Use the condition that the point $(2, -1)$ lies on the curve. Since the local maximum exists at $(2, -1)$, the point $(2, -1)$ must satisfy the function's equation. Substitute $x=2$ and $f(x)=-1$ into the given function $f(x) = \dfrac{ax+b}{(x-1)(x-4)}$:

Multiplying both sides by $-2$ gives our first equation:

2. Find the first derivative of $f(x)$. First, expand the denominator of $f(x)$:

We use the quotient rule for differentiation.
> [!FORMULA]
> If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.
Here, we have:
*   $u(x) = ax+b \implies u'(x) = a$
*   $v(x) = x^2-5x+4 \implies v'(x) = 2x-5$
Substituting these into the quotient rule formula:

3. Use the condition that the derivative is zero at the local maximum. At a local maximum, the first derivative $f'(x)$ must be zero. Since the local maximum is at $x=2$, we must have $f'(2)=0$. Substitute $x=2$ into the expression for $f'(x)$: …

We are told that $f(x) = \dfrac{ax+b}{(x-1)(x-4)}$ has a local maximum at the point $(2, -1)$. This gives us two pieces of information: the point lies on the graph, and the derivative is zero at $x = 2$. Let’s unpack why.

Concept and intuition:

A local maximum at a point means the function’s value there is higher than nearby values. For a differentiable function, this implies the tangent line is horizontal — so the derivative is zero. Also, the point must satisfy the function itself. So we have two equations in the unknowns $a$ and $b$. Solve them, then compute $a+b$.

1. Use the point on the curve Since $(2, -1)$ lies on $f(x)$, we have $f(2) = -1$.

Set equal to $-1$:

2. Use the derivative condition The derivative must be zero at $x = 2$. First, find $f'(x)$. Write $f(x) = \frac{ax+b}{x^2 - 5x + 4}$. Use the quotient rule:

We only need $f'(2) = 0$, so the numerator must be zero at $x = 2$ (denominator is nonzero there).

Numerator at $x = 2$:

Set equal to zero:

3. Solve for $a$ and $b$ …

Concept & Intuition

$f$ is decreasing where $f'(x)\le 0$ (with equality only at isolated points). We differentiate, simplify the sign, and take the longest interval on which $f$ never increases. The log term $\log\frac{1+x}{1-x}$ requires $\frac{1+x}{1-x}>0$, so the domain is $(-1,1)$; the answer must lie inside it.

Step-by-step

1. Domain. $\dfrac{1+x}{1-x}>0$ and $1-x^2\neq 0$ give $x\in(-1,1)$.

2. Differentiate.

The first two terms combine to $\dfrac{2}{1-x^2}$.

3. Last term (quotient rule).

4. Combine over $(1-x^2)^2$.

The numerator is

So

$f'(x)=\frac{-x^2(1+x^2)}{(1-x^2)^2}.$ …

A quadratic expression of the form $f(x) = Ax^2 + Bx + C$ represents a parabola when graphed. The sign of the coefficient $A$ determines the direction the parabola opens:

• If $A > 0$, the parabola opens upwards, and its vertex is the lowest point, representing the minimum value of the expression.
• If $A < 0$, the parabola opens downwards, and its vertex is the highest point, representing the maximum value of the expression.

In this problem, the expression is $x^2 + 5x - 2$. Here, the coefficient of $x^2$ is $A=1$, which is positive. Therefore, the parabola opens upwards, and the expression has a minimum value. This minimum value occurs at the vertex of the parabola.

The x-coordinate of the vertex for a quadratic $Ax^2 + Bx + C$ is given by a standard formula. Once we find this x-coordinate, which is $a$ in this problem, we can substitute it back into the expression to find the minimum value, $M$.

1. Identify the coefficients of the quadratic expression.

   The given quadratic expression is $x^2 + 5x - 2$.

   Comparing this to the standard form $Ax^2 + Bx + C$, we can identify the coefficients:

   $A = 1$

   $B = 5$

   $C = -2$

2. Find the x-coordinate where the minimum value exists.

   The problem states that the minimum value exists at $a$. This $a$ is the x-coordinate of the vertex of the parabola.

   For a quadratic expression $Ax^2 + Bx + C$, the x-coordinate of the vertex is given by $x = -\frac{B}{2A}$.

   Substitute the values of $A$ and $B$ into this formula:

   $a = -\frac{5}{2(1)}$

   $a = -\frac{5}{2}$

3. Find the minimum value of the expression.

   The problem states that the minimum value is $M$. This $M$ is the value of the expression when $x = a$.

   Substitute $a = -\frac{5}{2}$ into the expression $x^2 + 5x - 2$:

   $M = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) - 2$

   $M = \frac{25}{4} - \frac{25}{2} - 2$

   To combine these fractions, we find a common denominator, which is 4: …

The key idea: for a quadratic $ax^2 + bx + c$, the vertex (where the minimum or maximum occurs) is at $x = -\frac{b}{2a}$, and the value there is $f\left(-\frac{b}{2a}\right)$. Here the expression is $ax^2 - 7x + 3a$, so $b = -7$ and $c = 3a$. Since the minimum is given, $a$ must be positive (otherwise the parabola opens downward and has no minimum). We’ll use the vertex condition to find $a$, then use Vieta’s formulas to get the sum of the roots.

1. Find the vertex’s x-coordinate. For $f(x) = ax^2 - 7x + 3a$, the vertex is at

2. Compute the minimum value. Substitute $x = \frac{7}{2a}$ into $f(x)$:

Simplify term by term:

and the constant $+3a$ stays. So

3. Set this equal to the given minimum $-\frac{1}{8}$.

Multiply through by $8a$ (since $a > 0$, no sign issues):

Rearranging:

4. Solve for $a$. This quadratic in $a$ factors or use the quadratic formula:

Since $9409 = 97^2$ (check: $97^2 = 9409$), we have …

Ellipse. $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\Rightarrow a=5,\ c^2=25-16=9,\ c=3.$ Foci $(\pm3,0)$ and $e_1=\dfrac{3}{5}$, so $e_1^2=\dfrac{9}{25}.$

Hyperbola (same foci). $\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$ with $A^2+B^2=c^2=9$ and $e_2=\dfrac{3}{A}$, so $e_2^2=\dfrac{9}{A^2}.$

Condition. $e_2^2-e_1^2\ge 1$: …

The key idea is that for a fixed number of Bernoulli trials, the probability of exactly $k$ successes is given by the binomial formula. The "most probable" $k$ is the mode of this distribution. Instead of computing all probabilities, we can find where the sequence $P(k)$ stops increasing and starts decreasing — that is, where the ratio $P(k+1)/P(k)$ becomes less than 1.

1. Set up the binomial probability For $n=7$ trials, success probability $p = \frac{2}{5}$, failure probability $q = 1-p = \frac{3}{5}$, the probability of exactly $k$ successes is

2. Consider the ratio of successive probabilities

This ratio tells us how $P(k)$ changes as $k$ increases. If it is greater than 1, $P(k+1) > P(k)$; if less than 1, $P(k+1) < P(k)$.

3. Find where the ratio crosses 1 Set the ratio equal to 1 to find the threshold:

Solve:

4. Interpret the result The ratio is >1 when $k < 2.2$ (so probabilities increase up to $k=2$), and <1 when $k > 2.2$ (so probabilities decrease after $k=3$). This means the maximum occurs at the integer just after the crossing point: $k = 3$. …