Question

a) If the saving function is given by S° = 0.03Y^2 - 2Y + 150

Calculate the values of marginal propensity to save (MPS) and marginal propensity to

consume (MPC) when Y=40.


b) Find the second derivatives of these functions:

i) y=5x^2+350x-120 [3 marks]


ii) y= ln( x^2+4) [3 marks]



c) Find the relative maximum and minimum fory=3x^2+30x+605. [4 marks]

d) Find the absolute maximum and minimum for y=x^2-2x+3 between x=0 and x=3. [4 marks]

Answer 1

a) Marginal propensity to save (MPS)

The marginal propensity to save (MPS) function is expressed as the derivative of the savings (S) function with respect to disposable income (Y).

$$MPS=\dfrac{dS}{dY}=0.06Y-2$$

When $Y=40$

$$MPS=0.06(40)-2=0.4$$

$$1=MPC+MPS$$

$$MPC=1-MPS=1-(0.06Y-2)$$

$$=3-0.06Y$$

When $Y=40$

$$MPC=1-0.4=0.6$$

When $Y=40, MPS=0.4, MPC=0.6$

b) Find the second derivatives of these functions:

i)

$$y'=10x+350$$

$$y''=10$$

ii)

$$y'=\dfrac{2x}{x^2+4}$$

$$y''=2\cdot\dfrac{x^2+4-2x^2}{(x^2+4)^2}=-2\cdot\dfrac{x^2-4}{(x^2+4)^2}$$

c)

$$y'=6x+30$$

Critical number(s)

$$y'=0=>6x+30=0=>x=-5$$

Critical number: $-5$

$$y''=6>0$$

$y(-5)=3(-5)^2+30(-5)+605=530$

By the Second DerivativeTest the function $y$ has the relative minimum with the value of $530$ at $x=-5.$

The function $y$ has no the relative maximum.

d)

$$y'=2x-2$$

Critical number(s)

$$y'=0=>2x-2=0=>x=1$$

Critical number: $1$

$y(0)=(0)^2-2(0)+3=3$

$y(3)=(3)^2-2(3)+3=6$

$y(1)=(1)^2-2(1)+3=2$

$$2<3<6$$

The function $y$ has the absolute maximum with the value of $6$ at $x=3$ on $[0, 3].$

The function $y$ has the absolute minimum with the value of $2$ at $x=1$ on $[0, 3].$