solve the following mathematical problems:

Here, $p=3x^2-35x+50$

Now, if the company makes a profit, then

$$\begin{gathered} p>0 \\ \Rightarrow 3x^2-35x+50 >0 \\ \Rightarrow 3x^2-30x-5x+50 >0 \\ \Rightarrow 3x\left(x-10\right)-5\left(x-10\right) >0 \\ \Rightarrow \left(x-10\right)\left(3x-5\right) >0...............\left(i\right) \end{gathered}$$

As this equation (i) is greater than 0, so the value of the two roots must have different values in different intervals.

Now, from the equation (i), we have 

The value of x is greater than 10 ie: x> 10

Or, the value of x is less than $\frac{5}{3}$ and greater than or equal to 0 i.e. $0\le x<\frac{5}{3}$ 

Because advertising cost can not be negative.

So, if the company makes a profit,

The values of $x=\left\{0\le x\le \frac{5}{3} or, x>10\right\}$ . 

576 Tk. would be saved when the breadth = 3 meter less

1 Tk. would be saved when the breadth=$\frac{3}{576 }$  meter less

7,200 Tk. would be saved when the breadth$=\frac{3\times 7200}{576}=37.5$ meter less

So, the breadth of the room is 37.5 meter 

Since the sum of the 3-digits is II and each digit represents a prime number. 

So the 3-digits may be 2, 2, 7 or 3, 3, 5 because 2+2+7=11 and 3+3+5=11

Now using the digit 2, 2, 7 we have the prime number 227. Because other two numbers i.e. 722 and 272 are by 2 and thus are not prime. 

Again, using the digit 3, 3, 5 we have the number 353 which is a prime number. But 533 which is divisible by 13 and thus 533 is not prime and 335 is divisible by 5 and thus 335 is not prime.

So, the three-digit prime number whose sum of the digits is 11 and each digit representing a prime number is 227 & 353. (Answer)

To prove that a cyclic parallelogram must be a rectangle, we need to use the properties of cyclic quadrilaterals and parallelograms.

### Definitions and Properties:
1. **Cyclic Parallelogram**: A parallelogram is cyclic if all its vertices lie on a common circle.
2. **Parallelogram**: A parallelogram is a quadrilateral with opposite sides parallel and equal in length.
3. **Cyclic Quadrilateral**: A quadrilateral is cyclic if its vertices lie on a single circle. In a cyclic quadrilateral, opposite angles sum up to 180 degrees.

### Proof:

1. **Properties of Cyclic Quadrilaterals**:
  - For a cyclic quadrilateral, the sum of the opposite angles is \(180^\circ\). That is, if \(ABCD\) is a cyclic quadrilateral, then:
    \[
    \angle A + \angle C = 180^\circ
    \]
    \[
    \angle B + \angle D = 180^\circ
    \]

2. **Properties of a Parallelogram**:
  - In a parallelogram, opposite angles are equal. So, if \(ABCD\) is a parallelogram, then:
    \[
    \angle A = \angle C
    \]
    \[
    \angle B = \angle D
    \]

3. **Combining the Properties**:
  - Since \(ABCD\) is both a parallelogram and a cyclic quadrilateral, we use both sets of properties.

  - From the property of the cyclic quadrilateral, we have:
    \[
    \angle A + \angle C = 180^\circ
    \]

  - Since opposite angles in a parallelogram are equal, we also have:
    \[
    \angle A = \angle C
    \]

  - Substitute \(\angle C\) from the parallelogram property into the cyclic quadrilateral property:
    \[
    \angle A + \angle A = 180^\circ
    \]

    \[
    2 \angle A = 180^\circ
    \]

    \[
    \angle A = 90^\circ
    \]

  - Thus, each angle in the parallelogram \(ABCD\) is \(90^\circ\), which means all the angles in the parallelogram are right angles.

4. **Conclusion**:
  - Since a parallelogram with all angles equal to \(90^\circ\) is a rectangle, we conclude that a cyclic parallelogram must be a rectangle.

### Summary:
In a cyclic parallelogram, the property of the cyclic quadrilateral (opposite angles sum to \(180^\circ\)) combined with the property of the parallelogram (opposite angles are equal) shows that each angle in the parallelogram is \(90^\circ\). Therefore, the parallelogram must be a rectangle.

$$\begin{gathered} Let, \frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}=k \\ so, \frac{a}{q-r}=k \\ a=k\left(q-r\right);b=k\left(r-p\right);c=k\left(p-q\right) \\ L.H.S.=a+b+c \\ =k\left(q-r\right)+k\left(r-p\right)+k\left(p-q\right) \\ =k\left(q-r+r-p+p-q\right) \\ =k\times 0 \\ =0 \\ R.H.S = pa+qb+rc \\ =p\left[k\left(q-r\right)\right]+q\left[k\left(r-p\right)\right]+r\left[k\left(p-q\right)\right] \\ =p(kq-kr)+q(kr-kp)+r(kp-kq) \\ =pkq-pkr+qkr-pkq+pkr-qkr \\ =0 \end{gathered}$$

Therefore, a+b+c=pa+qb+ rc [Showed] 

Let,

The original speed = x km/h

ATQ, 

108/x - 108/x + 3 = 3

=> 108x + 324 - 108x/x² + 3x = 3

=> 3x² + 9x - 324 = 0

=> 3(x² + 3x - 108) = 0

=> x² + 3x - 108 = 0

=> x² + 12x - 9x - 108 = 0

=> (x + 12)(x - 9) = 0

=> x = 9

Therefore, original speed 9.

$$\begin{gathered} \frac{x}{2}+\frac{6}{y}=9 \\ \Rightarrow \frac{xy+12}{2y}=9 \\ xy+12=18y...........\left(i\right) \\ \frac{x}{3}+\frac{2}{y}=4 \\ \Rightarrow \frac{xy+6}{3y}=4 \\ xy+6=12y...........\left(ii\right) \\ Now, \left[\right(i)-(ii\left)\right] we have \\ \left(xy+12=18\right)-\left(xy+6=12y\right)=\left(6=6y\right)=\left(y=1 \right) \\ putting the value of \text{'}y\text{'} in \left(ii\right), we have \\ \left(x\times 1\right)+6=12\times 1 \\ \Rightarrow x=12-6 \\ x=6 \\ (x,y)=(6,1) \end{gathered}$$