a) Сonsider this figure:

From the figure it can be seen that:

Area \((A) = 4xy\)

And also,

\(x^{2}\ +\ 2y^{2} = 1\)

\(x = \sqrt{1\ -\ 2y^{2}\)

We've taken the positive value since we chose this point to be in the first quadrant

So now deciding:

\(A = 4xy\)

\(A = 4y \sqrt{1\ -\ 2y^{2}\)

Differentiating the above function with respect to "y":

\(\frac{dA}{dy}=\frac{d}{dy}\left(4y\sqrt{1\ -\ 2y^{2}}\right)\)

\(\frac{dA}{dy}=4y\frac{d}{dy}\sqrt{1\ -\ 4y^{2}}\ +\ \sqrt{1\ -\ 4y^{2}}\frac{d}{dy}(4y)\)

\(\frac{dA}{dy}=4y\frac{1}{2\sqrt{1\ -\ 4y^{2}}}(-8y)\ +\ 4\sqrt{1\ -\ 4y^{2}}\)

\(\frac{dA}{dy}=4\left[\frac{-8y^{2}\ +\ 2\ -\ 8y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]\)

\(\frac{dA}{dy}=4\left[\frac{2\ -\ 16y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]\)

For maximize the area:

Put,

\(\frac{dA}{dy} = 0\)

\(\frac{dA}{dy}=4\left[\frac{2\ -\ 16y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]\)

\(\left[\frac{2\ -\ 16y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]=0\)

\(2\ -\ 16y^{2} = 0\)

\(y^{2} = \frac{1}{8}\)

\(y = \frac{1}{\sqrt{8}}\)

Corresponding to this,

\(x=\sqrt{1\ -\ 2\ \times\ \frac{1}{8}}\)

\(x = \sqrt {1\ -\ \frac{1}{4}}\)

\(x = \sqrt{\frac{3}{4}}\)

\(x = \frac{\sqrt{3}}{2}}\)

Hence the maximum area:

Area \((A) = 4xy\)

\(Area_{max} = 4\ \times\ \frac{\sqrt{3}}{2}\ \times\ \frac{1}{\sqrt{8}}\)

\(Area_{max} = 2\ \times\ \frac{\sqrt{3}}{2 \sqrt{2}}\)

\(Area_{max} = \sqrt{\frac{3}{2}}\)

b)Prove the following: Let c in (a, b). If f is continuous on [a, b], differentiable on (a, b), and:

\(\lim_{x\ \rightarrow\ x}\ f'(x)=L\ then\ f'(c)=L\)

Properties used

\(\lim_{h\ \rightarrow\ 0}\ \frac{f'(x\ +\ h)\ -\ f(x)}{h}=f'(x)\)

Proof is given below:

Since:

\(\lim_{x\ \rightarrow\ c} f'(x) = L\)

By using the property:

\(\lim_{x\ \rightarrow\ c}\ \left[\lim_{h\ \rightarrow\ 0}\frac{f'(x\ +\ h)\ -\ f(x)}{h}\right]=L\)

\(\lim_{h\ \rightarrow\ 0}\ \left[\lim_{x\ \rightarrow\ c}\frac{f(x\ +\ h)\ -\ f(x)}{h}\right]=L\)

\([:' f(x) is continuous]\)

\(\lim_{h\ \rightarrow\ 0}\ \left[\frac{f(c\ +\ h)\ -\ f(c)}{h}\right]=L\)

\(\left[\lim_{h\ \rightarrow\ 0}\ \frac{f(c\ +\ h)\ -\ f(c)}{h}\right]=L\)

\(f' (c) = L\)

Answer:

a) \(Area_{max} = \sqrt{\frac{3}{2}}\)

b) The proof is given above.

From the figure it can be seen that:

Area \((A) = 4xy\)

And also,

\(x^{2}\ +\ 2y^{2} = 1\)

\(x = \sqrt{1\ -\ 2y^{2}\)

We've taken the positive value since we chose this point to be in the first quadrant

So now deciding:

\(A = 4xy\)

\(A = 4y \sqrt{1\ -\ 2y^{2}\)

Differentiating the above function with respect to "y":

\(\frac{dA}{dy}=\frac{d}{dy}\left(4y\sqrt{1\ -\ 2y^{2}}\right)\)

\(\frac{dA}{dy}=4y\frac{d}{dy}\sqrt{1\ -\ 4y^{2}}\ +\ \sqrt{1\ -\ 4y^{2}}\frac{d}{dy}(4y)\)

\(\frac{dA}{dy}=4y\frac{1}{2\sqrt{1\ -\ 4y^{2}}}(-8y)\ +\ 4\sqrt{1\ -\ 4y^{2}}\)

\(\frac{dA}{dy}=4\left[\frac{-8y^{2}\ +\ 2\ -\ 8y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]\)

\(\frac{dA}{dy}=4\left[\frac{2\ -\ 16y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]\)

For maximize the area:

Put,

\(\frac{dA}{dy} = 0\)

\(\frac{dA}{dy}=4\left[\frac{2\ -\ 16y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]\)

\(\left[\frac{2\ -\ 16y^{2}}{2\sqrt{1\ -\ 4y^{2}}}\right]=0\)

\(2\ -\ 16y^{2} = 0\)

\(y^{2} = \frac{1}{8}\)

\(y = \frac{1}{\sqrt{8}}\)

Corresponding to this,

\(x=\sqrt{1\ -\ 2\ \times\ \frac{1}{8}}\)

\(x = \sqrt {1\ -\ \frac{1}{4}}\)

\(x = \sqrt{\frac{3}{4}}\)

\(x = \frac{\sqrt{3}}{2}}\)

Hence the maximum area:

Area \((A) = 4xy\)

\(Area_{max} = 4\ \times\ \frac{\sqrt{3}}{2}\ \times\ \frac{1}{\sqrt{8}}\)

\(Area_{max} = 2\ \times\ \frac{\sqrt{3}}{2 \sqrt{2}}\)

\(Area_{max} = \sqrt{\frac{3}{2}}\)

b)Prove the following: Let c in (a, b). If f is continuous on [a, b], differentiable on (a, b), and:

\(\lim_{x\ \rightarrow\ x}\ f'(x)=L\ then\ f'(c)=L\)

Properties used

\(\lim_{h\ \rightarrow\ 0}\ \frac{f'(x\ +\ h)\ -\ f(x)}{h}=f'(x)\)

Proof is given below:

Since:

\(\lim_{x\ \rightarrow\ c} f'(x) = L\)

By using the property:

\(\lim_{x\ \rightarrow\ c}\ \left[\lim_{h\ \rightarrow\ 0}\frac{f'(x\ +\ h)\ -\ f(x)}{h}\right]=L\)

\(\lim_{h\ \rightarrow\ 0}\ \left[\lim_{x\ \rightarrow\ c}\frac{f(x\ +\ h)\ -\ f(x)}{h}\right]=L\)

\([:' f(x) is continuous]\)

\(\lim_{h\ \rightarrow\ 0}\ \left[\frac{f(c\ +\ h)\ -\ f(c)}{h}\right]=L\)

\(\left[\lim_{h\ \rightarrow\ 0}\ \frac{f(c\ +\ h)\ -\ f(c)}{h}\right]=L\)

\(f' (c) = L\)

Answer:

a) \(Area_{max} = \sqrt{\frac{3}{2}}\)

b) The proof is given above.