Prove these examples are correct: a) What is the area of the largest rectangle that fits inside of the ellipse x^{2} + 2y^{2} = 1? b) Prove the following: Let c in (a, b). If f is continuous on [a, b], differentiable on (a, b)?

Question
Transformation properties
asked 2021-02-25
Prove these examples are correct:
a) What is the area of the largest rectangle that fits inside of the ellipse
\(x^{2}\ +\ 2y^{2} = 1?\)
b) Prove the following: Let c in (a, b). If f is continuous on \([a,\ b],\) differentiable on (a, b)?

Answers (1)

2021-02-26
a) Сonsider this figure:
image
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.
0

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