Calculus Examples
D(p)=1500-12p , p=100
Step 1
Write D(p)=1500-12p as an equation.
q=1500-12p
Step 2
To find elasticity of demand, use the formula E=|pqdqdp|.
Step 3
Step 3.1
Substitute 100 for p.
q=1500-12⋅100
Step 3.2
Multiply -12 by 100.
q=1500-1200
Step 3.3
Subtract 1200 from 1500.
q=300
q=300
Step 4
Step 4.1
Differentiate the demand function.
dqdp=ddp[1500-12p]
Step 4.2
Differentiate.
Step 4.2.1
By the Sum Rule, the derivative of 1500-12p with respect to p is ddp[1500]+ddp[-12p].
dqdp=ddp[1500]+ddp[-12p]
Step 4.2.2
Since 1500 is constant with respect to p, the derivative of 1500 with respect to p is 0.
dqdp=0+ddp[-12p]
dqdp=0+ddp[-12p]
Step 4.3
Evaluate ddp[-12p].
Step 4.3.1
Since -12 is constant with respect to p, the derivative of -12p with respect to p is -12ddp[p].
dqdp=0-12ddp[p]
Step 4.3.2
Differentiate using the Power Rule which states that ddp[pn] is npn-1 where n=1.
dqdp=0-12⋅1
Step 4.3.3
Multiply -12 by 1.
dqdp=0-12
dqdp=0-12
Step 4.4
Subtract 12 from 0.
dqdp=-12
dqdp=-12
Step 5
Step 5.1
Substitute -12 for dqdp.
E=|pq⋅-12|
Step 5.2
Substitute the values of p and q.
E=|100300⋅-12|
Step 5.3
Cancel the common factor of 12.
Step 5.3.1
Factor 12 out of 300.
E=|10012(25)⋅-12|
Step 5.3.2
Factor 12 out of -12.
E=|10012⋅25⋅(12⋅-1)|
Step 5.3.3
Cancel the common factor.
E=|10012⋅25⋅(12⋅-1)|
Step 5.3.4
Rewrite the expression.
E=|10025⋅-1|
E=|10025⋅-1|
Step 5.4
Combine 10025 and -1.
E=|100⋅-125|
Step 5.5
Multiply 100 by -1.
E=|-10025|
Step 5.6
Divide -100 by 25.
E=|-4|
Step 5.7
The absolute value is the distance between a number and zero. The distance between -4 and 0 is 4.
E=4
E=4
Step 6
Since E>1, the demand is elastic.
E=4
Elastic