Topic : Simultaneous Equation

Problem : Solve 3x -2y = 3 and x = y + 4

Solution :

3x -2y = 3 ——- (1)

x = y + 4 ——- (2)

Use equation (2) in (1) for substitution method

So, 3x -2y = 3

3(y+4) – 2y = 3

3y + 12 – 2y = 3

y + 12 = 3

Subtract 12 from both sides

y = – 9

Substitute y = -9 in equation (2)

x = y + 4

x = -9 + 4 

x = - 5

Topic : Integration

Question : Integrate e^1/x / x²

Solution :  

Topic : Integration

Question : Integrate   ∫ (2+3Sin³ x)/ Sin² x dx

Solution :

Topic : Straight Line Equation

Question :  Find the equation of the straight line with slope 1/3 and passes through the point ( 3, -3)

Answer :

Given point is (3,-3)

and slope 1/3

THe slope intercept form of a line is y = mx + b;

m is slope and

b is y intercept

Given y=m/3 so we have y= (1/3)x + b

This line passess through (3,-3), So we substitute x=3 and y = -3

y=(1/3)x+b

-3=(1/3)*3+b

-3=1+b

-1=-1  subtracting 1 on both the sides

——-

-4=b

Hence substitute b = -4

So the required equstion is y = (1/3)x -4

we can simplify further as,

3y= 3[(1/3)x-4]

3y= 3*(1/3)x -3*4

3y=x-12 is the equation of straight line.

Topic : Properties of Triangle

Question : List and Explain the properties of Triangle.

Solution :

Vertex : The vertex (plural: vertices) is a corner of the triangle. Every triangle has three vertices. 

Base : The base of a triangle can be any one of the three sides, usually the one drawn at the bottom. You can pick any side you like to be the base. Commonly used as a reference side for calculating the area of the triangle. In an isosceles triangle, the base is usually taken to be the unequal side. 

Altitude : The altitude of a triangle is the perpendicular from the base to the opposite vertex. The altitudes intersect at a single point, called the orthocenter of the triangle.

Median : The median of a triangle is a line from a vertex to the midpoint of the opposite side. 

Area : The number of square units it takes to exactly fill the interior of a triangle.

Perimeter : The distance around the triangle. The sum of its sides. Definition: The total distance around the outside of a triangle

perimeter = a+b+c, where a,b and c are the lengths of each side of the triangle

Topic : Word Problem on Determining the Revenue Function.

Question : A cable television firm presently serves 5000 households and charges $20 per month. A marketing survey indicates that each decrease of $1 in the monthly charge will result in 500 new customers. Let R(x) denote the total monthly revenue when the monthly charge is x dollars. Determine the revenue function R. find the value of x that results in maximum monthly revenue.

Answer :

Total monthly revenue = number of customers x monthly rent
for $20 rent : Total monthly revenue = 5000 x 20 = $100000
for every $1 decrease in the rent , there is an increase of 500 customers.
So , the relation is formulated as follows:

decrease of $1 : monthly rent = $(20-1) = $19
new customers = 500 = 500 x 1 ,
total number of customers = 5000 + 500 = 5500 = 5000 + 500 x 1
total monthly revenue = (5000 + 500 x 1) x (20 – 1)

decrease of $2 : monthly rent = $(20-2) = $18
new customers = 500 + 500 = 1000 = 500 x 2 ,
total number of customers = 5000 + 1000 = 6000 = 5000 + 500 x 2
total monthly revenue = (5000 + 500 x 2) x (20 – 1)

decrease of $3 : monthly rent = $(20-3) = $17
new customers = 500 + 500 + 500 = 1500 ,
total number of customers = 5000 + 1500 = 6500 = 5000 + 500 x 3
total monthly revenue = (5000 + 500 x 3) x (20 – 3)
……………………..
the relation is now : R(x) = (5000 + 500 x X) x ( 20 – x) = 500 (10 + X)(20-X)
to find the max. monthly revenue we have to find the first derivative of R(x) and equate it to 0.
d/dx ( R(x)) = 500 { (10 + X)(-1) + (1)(20 – X) } = 500 { 10 – 2X }
we should equate that to 0 .
so, 10 – 2x = 0 , so , X = 5
So , the total monthly revenue will be maximum if the monthly rent is decreased by $5.
Finally the answer is :
Revenue Function is : R(x) = (5000 + 500 x X) x ( 20 – x)
The total monthly revenue will be maximum if the monthly rent is decreased by $5 ; i.e if monthly rent is $15 the total monthly revenue will be maximum.