question_answer

Prove that the surface area of a solid cuboid of square base and given volume is minimum when it is a cube.

OR

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is \[(\pi /3).\]

Answer:

Let a be the side of the square base of the cuboid and h be its height. Let S be its surface area and V be the volume. Then,

\[V={{a}^{2}}h\]                       [given] ...(i)

and       \[S=2{{a}^{2}}+4ah=2{{a}^{2}}+4a\cdot \frac{V}{{{a}^{2}}}\]

[from Eq. (i)]

\[\Rightarrow \]   \[S=2{{a}^{2}}+\frac{4V}{a}\]                                   ? (ii)

On differentiating both sides twice w.r.t. a, we get

\[\frac{dS}{da}=4a-\frac{4V}{{{a}^{2}}}\]

and       \[\frac{{{d}^{2}}S}{d{{a}^{2}}}=4+\frac{8V}{{{a}^{3}}}\]

For maximum or minimum, put \[\frac{dS}{da}=0.\]

\[\Rightarrow \]   \[4a-\frac{4V}{{{a}^{2}}}=0\] \[\Rightarrow \] \[{{a}^{3}}=V\] \[\Rightarrow \] \[a={{V}^{1/3}}\]

At \[a={{V}^{1/3}},\] \[\frac{{{d}^{2}}S}{d{{a}^{2}}}=4+\frac{8V}{V}=12>0\]

\[\Rightarrow \] S is minimum, when \[a={{V}^{1/3}}.\]

On putting \[V={{a}^{3}}\] in Eq. (i), we get

\[{{a}^{3}}={{a}^{2}}h\] \[\Rightarrow \] \[h=\frac{{{a}^{3}}}{{{a}^{2}}}=a\]

\[\Rightarrow \] All sides of the cuboid are equal.

Hence, in this case the cuboid is a cube.

Hence proved,

OR

Let us consider a right-angled triangle with base = x and hypotenuse = y. Let \[x+y=k,\] where k is a constant. Let \[\theta \] be the angle between the base and the hypotenuse. Let A be the area of the triangle. Then,

\[A=\frac{1}{2}\times BC\times AC=\frac{1}{2}x\sqrt{{{y}^{2}}-{{x}^{2}}}.\]

\[\therefore \]      \[{{A}^{2}}=\frac{{{x}^{2}}({{y}^{2}}-{{x}^{2}})}{4}=\frac{{{x}^{2}}[{{(k-x)}^{2}}-{{x}^{2}}]}{4}\]    \[[\because \,\,\,\,y=(k-x)]\]

or         \[{{A}^{2}}==\frac{{{k}^{2}}{{x}^{2}}-2k{{x}^{3}}}{4}\]                      ?(i)

On differentiating (i), we get

\[2A\cdot \frac{dA}{dx}=\frac{2{{k}^{2}}x-6k{{x}^{2}}}{4}\]                        ? (ii)

or         \[\frac{dA}{dx}=\frac{{{k}^{2}}x-3k{{x}^{2}}}{4A}\]

Now, \[\frac{dA}{dx}=0\] \[\Rightarrow \] \[({{k}^{2}}x-3k{{x}^{2}})=0\] \[\Rightarrow \] \[x=\frac{k}{3}\]

[Neglecting x = 0].

Now, differentiating (ii), we get

\[2{{\left( \frac{dA}{dx} \right)}^{2}}+2A\cdot \frac{{{d}^{2}}A}{d{{x}^{2}}}=\frac{2{{k}^{2}}-12kx}{4}\]        ? (iii)

Putting \[\frac{dA}{dx}=0\] and \[x=\frac{k}{3}\] in (iii), we get \[\frac{{{d}^{2}}A}{d{{x}^{2}}}=\frac{-{{k}^{2}}}{4A}<0.\]

Thus, A is maximum when \[x=\left( \frac{k}{3} \right).\]

Now,     \[x=\frac{k}{3}\] \[\Rightarrow \] \[y=\left( k-\frac{k}{3} \right)=\frac{2k}{3}.\]

\[\therefore \]   \[\frac{x}{y}=\cos \theta \] \[\Rightarrow \] \[\cos \theta =\frac{(k\,/3)}{(2k\,/3)}=\frac{1}{2}\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}\]

Hence, the area is maximum when \[\theta =\frac{\pi }{3}.\]