tg

tg
tgt

Sunday, 3 August 2014

CHAPTER 16 - Worked Out Examples

Example: 1      

Find the extrema points of f\left( x \right) = 3{x^4} - 4{x^3} - 36{x^2} + 28.
Solution: 1

f'\left( x \right) = 12{x^3} - 12{x^2} - 72x
 = 12x\left( {{x^2} - x + 6} \right)
 = 12x\left( {{x^2} - x + 6} \right)
We determine the sign of f'(x) using a number line:
From the number line, observe that (using the FODT):
x =  - 2\,{\rm{and }}\,x = 3 are local minima
x = 0 is a local maximum
Alternatively, we can use the SODT:
f''\left( x \right) = 36{x^2} - 24x - 72
 = 12\left( {3{x^2} - 2x - 6} \right)
f''\left( 0 \right) < 0 \,\,\,\,  \Rightarrow  \,\,\,\, x = 0 is a local maximum
f''\left( { - 2} \right) > 0 \,\,\,\,  \Rightarrow  \,\,\,\, x =  - 2 is a local minimum
f''\left( 3 \right) > 0 \,\,\,\,  \Rightarrow  \,\,\,\, x = 3 is a local minimum
     Example: 2 
Let f\left( x \right) = 2{x^3} - 3\left( {a + b} \right){x^2} + 6abx. If a < b, determine the local maximum/minimum points of f(x). If a = b, how will the answer change?
Solution: 2

f'(x) = 6{x^2} - 6(a + b)x + 6ab
 = 6(x - a)(x - b)
To determine the sign of f'(x) in different intervals, we use a number line:
f'(x) > 0\,\,:\,\,x < a or x>b
f'(x) < 0\,\,:\,\,a < x < b
Observe that f'(x) changes from positive to negative in the neighbourhood of x = a.
 \Rightarrow\,\,\,\, x = a is a point of local maximum
Similarly, f'(x) changes from negative to positive in the neighbourhood of x = b.
 \Rightarrow\,\,\,\, x = b is a point of local minimum
If a = b,f'(x) = 6{(x - a)^2}
Notice that f'(x) is never negative. f'(x) is always positive except at x = awhere f'(x) = 0
 \Rightarrow\,\,\,\, x = a is a point of inflexion
     Example: 3     

Let (h,k) be a fixed point, where h > 0,k > 0. A straight line passing through this point cuts the positive direction of the co-ordinate axes at the points P and Q. Find the minimum area of \Delta OPQO being the origin.
Solution: 3

The given point (h,k) will lie in the first quadrant.
Convince yourself that there will be a particular slope of PQ at which the area of \Delta OPQ is minimum. If the \left( {{\rm{slope}}} \right) \to 0 or \left( {{\rm{slope}}} \right) \to \infty \left( {{\rm{area}}} \right) \to \infty . However, at some finite slope in between these two extremes, area will assume a minimum value.
Assume m to be the slope we wish to determine so that area is minimum (Notice that m will be negative). We first write down the equation of a straight line passing through (h,k) with slope m:
y - k = m(x - h)
This cuts the X-axis at P\left( {h - \dfrac{k}{m},0} \right) and the Y-axis at Q(0,k - mh)
Assume A to be the area of \Delta OPQ.
Therefore,
A = \dfrac{1}{2} \times OP \times OQ
 = \dfrac{1}{2} \times \left( {h - \dfrac{k}{m}} \right) \times \left( {k - mh} \right)
 = \dfrac{{ - 1}}{{2m}}{\left( {k - mh} \right)^2}\ldots(i)
For A to be minimum,
\dfrac{{dA}}{{dm}} = 0\,\,\,{\rm{and}}\;\,\dfrac{{{d^2}A}}{{{d^2}m}} > 0
 \Rightarrow  \,\,\,\, \dfrac{{dA}}{{dm}} = \dfrac{{ - 1}}{{2m}}.2\left( {k - mh} \right).\left( { - h} \right) + {\left( {k - mh} \right)^2}.\dfrac{1}{{2{m^2}}}
 = \dfrac{1}{{2{m^2}}}\left( {{k^2} - {m^2}{h^2}} \right)
 \Rightarrow  \,\,\,\, \dfrac{{dA}}{{dm}} = 0\,\,\,{\rm{when  }}{k^2} - {m^2}{h^2} = 0
 \Rightarrow  \,\,\,\, m =  \pm \dfrac{k}{h}
Since m must be negative, m = \dfrac{{ - k}}{h}.
Now,
{\left. {\dfrac{{{d^2}A}}{{d{m^2}}}} \right|_{m = \dfrac{{ - k}}{h}}}{\left. { = \dfrac{1}{{2{m^2}}}\left( { - 2m{h^2}} \right) - \dfrac{{\left( {{k^2} - {m^2}{h^2}} \right)}}{{{m^3}}}} \right|_{m = \dfrac{{ - k}}{h}}}
 = \dfrac{{{h^3}}}{k} > 0
 \Rightarrow \,\,\,\,m = \dfrac{{ - k}}{h} is a point of local minimum for A
From (i), the minimum value of A is :
{\left. {{A_{\min }} = \dfrac{{ - 1}}{{2m}}{{\left( {k - mh} \right)}^2}} \right|_{m = \dfrac{{ - k}}{h}}} = 2hk
Post a Comment