1 Determine the absolute max and min of f(x)=x^4-2x^2+5 on the interval -2<x<1.

Question

1 Determine the absolute max and min of f(x)=x^4-2x^2+5 on the interval -2<x<1.

2. The total cost of producing x units of a certain commodity is C(x)=x^3-5x^2+8x. Determine the minimum average cost of the commodity. How many units are needed to produce the min average cost?

3. Suppose that the demand equation for a certain commidity is q=200-2p^2;0<p<10. Express the elasticity of demand as a function of p. Calculate the elasticity when p=6. Interpret answer in a sentence. At what price is the elasticity of demand equal to 1?

Explanation / Answer

1) f(x)=x^4-2x^2+5

f'(x)= 4x^3-4x

to find min and max ,f'(x)=0

4x^3-4x =0

x=0 and x= 1, -1

now for absolute min and max we need to check the value of f(x) at x= 0 ,1 , -1 and -2

f(0)=5

f(1)=f(-1)=4

f(-2)=13

so the absolute max in the interval is 13

and the absolute minimum in the interval is 4

2. average cost function = A(x)=C(x)/x= x^2-5x+8

For minimum A(x), A'(x)=0

2x-5=0

x=2.5   Units to produce the minimum average cost

A(2.5)=(2.5)^2-5(2.5)+8= 6.25+8-12.5=1.75    minimum average cost

3) elasticity = (p/q)*dq/dp

dq/dp=-4p

so elasticity = (p/(200-2p^2)) * (-4p)= 4p^2/(2p^2-200)= 2p^2/(p^2-100)

Elasticity at (p=6) = 2*36/(36-100)= -72/64 = -9/8     Elastic or relatively elastic demand.

2p^2/(p^2-100)= 1

2p^2=p^2-100

p^2= -100

So no real value of price can give us elasticity 1

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