(3) Find the maxima and minima of
y
=
x
−
1
x
2
+
2
{\displaystyle y={\dfrac {x-1}{x^{2}+2}}}
d
y
d
x
=
(
x
2
+
2
)
×
1
−
(
x
−
1
)
×
2
x
(
x
2
+
2
)
2
=
2
x
−
x
2
+
2
(
x
2
+
2
)
2
=
0
{\displaystyle {\frac {dy}{dx}}={\frac {(x^{2}+2)\times 1-(x-1)\times 2x}{(x^{2}+2)^{2}}}={\frac {2x-x^{2}+2}{(x^{2}+2)^{2}}}=0}
or
x
2
−
2
x
−
2
=
0
{\displaystyle x^{2}-2x-2=0}
x
=
+
2.73
{\displaystyle x=+2.73}
x
=
−
0.73
{\displaystyle x=-0.73}
d
2
y
d
x
2
=
−
(
x
2
+
2
)
2
×
(
2
x
−
2
)
−
(
x
2
−
2
x
−
2
)
(
4
x
3
+
8
x
)
(
x
2
+
2
)
4
=
−
2
x
5
−
6
x
4
−
8
x
3
−
8
x
2
−
24
x
+
8
(
x
2
+
2
)
4
{\displaystyle {\begin{aligned}{\dfrac {d^{2}y}{dx^{2}}}&=-{\frac {(x^{2}+2)^{2}\times (2x-2)-(x^{2}-2x-2)(4x^{3}+8x)}{(x^{2}+2)^{4}}}\\&=-{\frac {2x^{5}-6x^{4}-8x^{3}-8x^{2}-24x+8}{(x^{2}+2)^{4}}}\end{aligned}}}
The denominator is always positive, so it is sufficient to ascertain the sign of the numerator.
If we put
x
=
2.73
{\displaystyle x=2.73}
y
=
0.183
{\displaystyle y=0.183}
If we put
x
=
−
0.73
{\displaystyle x=-0.73}
y
=
−
0.683
{\displaystyle y=-0.683}
(4) The expense
C
{\displaystyle C}
P
{\displaystyle P}
C
=
a
P
+
b
c
+
P
+
d
{\displaystyle C=aP+{\dfrac {b}{c+P}}+d}
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
d
{\displaystyle d}
d
C
d
P
=
a
−
b
(
c
+
P
)
2
=
0
{\displaystyle {\dfrac {dC}{dP}}=a-{\frac {b}{(c+P)^{2}}}=0\quad }
hence
a
=
b
(
c
+
P
)
2
{\displaystyle a={\dfrac {b}{(c+P)^{2}}}}
P
=
±
b
a
−
c
{\displaystyle P=\pm {\sqrt {\dfrac {b}{a}}}-c}
As the output cannot be negative,
P
=
+
b
a
−
c
{\displaystyle P=+{\sqrt {\dfrac {b}{a}}}-c}
