You will get
d
S
d
x
=
x
×
−
x
4
R
2
−
x
2
+
4
R
2
−
x
2
=
4
R
2
−
2
x
2
4
R
2
−
x
2
{\displaystyle {\dfrac {dS}{dx}}=x\times -{\dfrac {x}{\sqrt {4R^{2}-x^{2}}}}+{\sqrt {4R^{2}-x^{2}}}={\dfrac {4R^{2}-2x^{2}}{\sqrt {4R^{2}-x^{2}}}}}
.
For maximum or minimum we must have
4
R
2
−
2
x
2
4
R
2
−
x
2
=
0
{\displaystyle {\dfrac {4R^{2}-2x^{2}}{\sqrt {4R^{2}-x^{2}}}}=0}
;
that is,
4
R
2
−
2
x
2
=
0
{\displaystyle 4R^{2}-2x^{2}=0}
and
x
=
R
2
{\displaystyle x=R{\sqrt {2}}}
.
The other side
=
4
R
2
−
2
R
2
=
R
2
{\displaystyle ={\sqrt {4R^{2}-2R^{2}}}=R{\sqrt {2}}}
; the two sides are equal; the figure is a square the side of which is equal to the diagonal of the square constructed on the radius. In this case it is, of course, a maximum with which we are dealing.
(2) What is the radius of the opening of a conical vessel the sloping side of which has a length
l
{\displaystyle l}
when the capacity of the vessel is greatest?
If
R
{\displaystyle R}
be the radius and
H
{\displaystyle H}
the corresponding height,
H
=
l
2
−
R
2
{\displaystyle H={\sqrt {l^{2}-R^{2}}}}
.
Volume
V
=
π
R
2
×
H
3
=
π
R
2
×
l
2
−
R
2
3
{\displaystyle V=\pi R^{2}\times {\dfrac {H}{3}}=\pi R^{2}\times {\dfrac {\sqrt {l^{2}-R^{2}}}{3}}}
.
Proceeding as in the previous problem, we get
d
V
d
R
=
π
R
2
×
−
R
3
l
2
−
R
2
+
2
π
R
3
l
2
−
R
2
{\displaystyle {\dfrac {dV}{dR}}=\pi R^{2}\times -{\dfrac {R}{3{\sqrt {l^{2}-R^{2}}}}}+{\dfrac {2\pi R}{3}}{\sqrt {l^{2}-R^{2}}}}
=
2
π
R
(
l
2
−
R
2
)
−
π
R
3
3
l
2
−
R
2
=
0
{\displaystyle ={\dfrac {2\pi R(l^{2}-R^{2})-\pi R^{3}}{3{\sqrt {l^{2}-R^{2}}}}}=0}
for maximum or minimum.
Or,
2
π
R
(
l
2
−
R
2
)
−
π
R
2
=
0
{\displaystyle 2\pi R(l^{2}-R^{2})-\pi R^{2}=0}
, and
R
=
l
2
3
{\displaystyle R=l{\sqrt {\tfrac {2}{3}}}}
for a maximum, obviously.