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Application of Derivatives

85737 Let \(A D\) and \(B C\) be two vertical poles at \(A\) and \(B\) respectively on a horizontal ground, If \(A D=8 \mathrm{~m}\), \(B C=11 \mathrm{~m}\) and \(A B=10 \mathrm{~m}\), then the distance (in meters) of point \(M\) and \(A B\) from the point \(A\) such that \(\mathrm{MD}^{2}+\mathrm{MC}^{2}\) is minimum is

1 8

2 5

3 4

4 7

MHT CET-2022

Application of Derivatives

85738 A rectangle of maximum area is inscribed in an ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\), then its dimensions are

1 \(4 \sqrt{2}, 5 \sqrt{2}\) (units)

2 \(\sqrt{2}, 5 \sqrt{2}\) (units)

3 \(4 \sqrt{2}, \sqrt{2}\) (units)

4 \(4 \sqrt{2}, 6 \sqrt{2}\) (units)

Explanation:

(A) : Given the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Let, there is a point \((\mathrm{h}, \mathrm{k})\) on the ellipse. So, we get \(\frac{\mathrm{h}^{2}}{25}+\frac{\mathrm{k}^{2}}{16}=1\)
\(\therefore \frac{\mathrm{h}^{2}}{25}=1-\left(\frac{\mathrm{k}}{4}\right)^{2} \Rightarrow \mathrm{h}=5 \sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\)

Area of the rectangle A \(=2 \mathrm{~h} \times 2 \mathrm{k}\)
\(=4 \mathrm{hk} \text {. }\)
Putting the value of \(h\). we get -
\(A=4 \times 5 \sqrt{1-\left(\frac{k}{4}\right)^{2}} \cdot \mathrm{k}\)
Now, differentiating w.r.t. k, we get -
\(\frac{\mathrm{dA}}{\mathrm{dk}}=20 \cdot \mathrm{k} \cdot\left[\frac{1}{2}\left[1-\left(\frac{\mathrm{k}}{4}\right)^{2}\right]^{-1 / 2}\left[0-\frac{2 \mathrm{k}}{16}\right]\right]+\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(=(20)\left[-\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}}+\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\right]\)
For max. k,
\(\frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}}=\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(\Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=1-\left(\frac{\mathrm{k}}{4}\right)^{2}\)
\(\Rightarrow 2\left(\frac{\mathrm{k}}{4}\right)^{2}=1 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=\frac{1}{2}\)
\(\therefore \mathrm{k}=\frac{4}{\sqrt{2}}=2 \sqrt{2}\)
So, we get
\(\frac{\mathrm{h}^{2}}{25}=1-\frac{\mathrm{k}^{2}}{16}=1-\frac{8}{16}=\frac{1}{2}\)
\(\therefore \quad \mathrm{h}^{2}=\frac{25}{2}\)
Or \(\quad \mathrm{h}=\frac{5}{\sqrt{2}}\)
Dimensions of the rectangle are
\(\begin{array}{ll}\therefore \quad 2 \mathrm{k} =2 \times 2 \sqrt{2}=4 \sqrt{2} \\ \quad 2 \mathrm{~h} =5 \sqrt{2}\end{array}\)
\(2 h=5 \sqrt{2}\)
So, the dimensions are \((4 \sqrt{2}, 5 \sqrt{2})\) units.

MHT CET-2021

Application of Derivatives

85739 A wire of length 20 units is divided into two parts such that the product of one part and cube of the other part is maximum, then product of these parts is

1 5 units

2 70 units

3 75 units

4 15 units

MHT CET-2021

Application of Derivatives

85740 Let \(f(x)=x^{2} \log x, x>0\). Then the minimum value of \(f\) is

1 \(\frac{1}{\sqrt{\mathrm{e}}}\)

2 \(2 \mathrm{e}\)

3 \(-2 \mathrm{e}\)

4 \(\sqrt{\mathrm{e}}\)

5 \(\frac{-1}{2 \mathrm{e}}\)

Kerala CEE-2021

Application of Derivatives

85741 The length of the longest size rectangle of maximum area that can be inscribed in a semicircle of radius 1 , so that 2 vertices lie on the diameter, is:

1 \(\sqrt{2}\)

2 2

3 \(\sqrt{3}\)

4 \(\frac{\sqrt{2}}{3}\)

5 \(\frac{-2}{\sqrt{3}}\)

Kerala CEE-2004

Application of Derivatives

85737 Let \(A D\) and \(B C\) be two vertical poles at \(A\) and \(B\) respectively on a horizontal ground, If \(A D=8 \mathrm{~m}\), \(B C=11 \mathrm{~m}\) and \(A B=10 \mathrm{~m}\), then the distance (in meters) of point \(M\) and \(A B\) from the point \(A\) such that \(\mathrm{MD}^{2}+\mathrm{MC}^{2}\) is minimum is

1 8

2 5

3 4

4 7

MHT CET-2022

Application of Derivatives

85738 A rectangle of maximum area is inscribed in an ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\), then its dimensions are

1 \(4 \sqrt{2}, 5 \sqrt{2}\) (units)

2 \(\sqrt{2}, 5 \sqrt{2}\) (units)

3 \(4 \sqrt{2}, \sqrt{2}\) (units)

4 \(4 \sqrt{2}, 6 \sqrt{2}\) (units)

Explanation:

(A) : Given the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Let, there is a point \((\mathrm{h}, \mathrm{k})\) on the ellipse. So, we get \(\frac{\mathrm{h}^{2}}{25}+\frac{\mathrm{k}^{2}}{16}=1\)
\(\therefore \frac{\mathrm{h}^{2}}{25}=1-\left(\frac{\mathrm{k}}{4}\right)^{2} \Rightarrow \mathrm{h}=5 \sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\)

Area of the rectangle A \(=2 \mathrm{~h} \times 2 \mathrm{k}\)
\(=4 \mathrm{hk} \text {. }\)
Putting the value of \(h\). we get -
\(A=4 \times 5 \sqrt{1-\left(\frac{k}{4}\right)^{2}} \cdot \mathrm{k}\)
Now, differentiating w.r.t. k, we get -
\(\frac{\mathrm{dA}}{\mathrm{dk}}=20 \cdot \mathrm{k} \cdot\left[\frac{1}{2}\left[1-\left(\frac{\mathrm{k}}{4}\right)^{2}\right]^{-1 / 2}\left[0-\frac{2 \mathrm{k}}{16}\right]\right]+\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(=(20)\left[-\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}}+\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\right]\)
For max. k,
\(\frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}}=\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(\Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=1-\left(\frac{\mathrm{k}}{4}\right)^{2}\)
\(\Rightarrow 2\left(\frac{\mathrm{k}}{4}\right)^{2}=1 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=\frac{1}{2}\)
\(\therefore \mathrm{k}=\frac{4}{\sqrt{2}}=2 \sqrt{2}\)
So, we get
\(\frac{\mathrm{h}^{2}}{25}=1-\frac{\mathrm{k}^{2}}{16}=1-\frac{8}{16}=\frac{1}{2}\)
\(\therefore \quad \mathrm{h}^{2}=\frac{25}{2}\)
Or \(\quad \mathrm{h}=\frac{5}{\sqrt{2}}\)
Dimensions of the rectangle are
\(\begin{array}{ll}\therefore \quad 2 \mathrm{k} =2 \times 2 \sqrt{2}=4 \sqrt{2} \\ \quad 2 \mathrm{~h} =5 \sqrt{2}\end{array}\)
\(2 h=5 \sqrt{2}\)
So, the dimensions are \((4 \sqrt{2}, 5 \sqrt{2})\) units.

MHT CET-2021

Application of Derivatives

85739 A wire of length 20 units is divided into two parts such that the product of one part and cube of the other part is maximum, then product of these parts is

1 5 units

2 70 units

3 75 units

4 15 units

MHT CET-2021

Application of Derivatives

85740 Let \(f(x)=x^{2} \log x, x>0\). Then the minimum value of \(f\) is

1 \(\frac{1}{\sqrt{\mathrm{e}}}\)

2 \(2 \mathrm{e}\)

3 \(-2 \mathrm{e}\)

4 \(\sqrt{\mathrm{e}}\)

5 \(\frac{-1}{2 \mathrm{e}}\)

Kerala CEE-2021

Application of Derivatives

85741 The length of the longest size rectangle of maximum area that can be inscribed in a semicircle of radius 1 , so that 2 vertices lie on the diameter, is:

1 \(\sqrt{2}\)

2 2

3 \(\sqrt{3}\)

4 \(\frac{\sqrt{2}}{3}\)

5 \(\frac{-2}{\sqrt{3}}\)

Kerala CEE-2004

Application of Derivatives

85737 Let \(A D\) and \(B C\) be two vertical poles at \(A\) and \(B\) respectively on a horizontal ground, If \(A D=8 \mathrm{~m}\), \(B C=11 \mathrm{~m}\) and \(A B=10 \mathrm{~m}\), then the distance (in meters) of point \(M\) and \(A B\) from the point \(A\) such that \(\mathrm{MD}^{2}+\mathrm{MC}^{2}\) is minimum is

1 8

2 5

3 4

4 7

MHT CET-2022

Application of Derivatives

85738 A rectangle of maximum area is inscribed in an ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\), then its dimensions are

1 \(4 \sqrt{2}, 5 \sqrt{2}\) (units)

2 \(\sqrt{2}, 5 \sqrt{2}\) (units)

3 \(4 \sqrt{2}, \sqrt{2}\) (units)

4 \(4 \sqrt{2}, 6 \sqrt{2}\) (units)

Explanation:

(A) : Given the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Let, there is a point \((\mathrm{h}, \mathrm{k})\) on the ellipse. So, we get \(\frac{\mathrm{h}^{2}}{25}+\frac{\mathrm{k}^{2}}{16}=1\)
\(\therefore \frac{\mathrm{h}^{2}}{25}=1-\left(\frac{\mathrm{k}}{4}\right)^{2} \Rightarrow \mathrm{h}=5 \sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\)

Area of the rectangle A \(=2 \mathrm{~h} \times 2 \mathrm{k}\)
\(=4 \mathrm{hk} \text {. }\)
Putting the value of \(h\). we get -
\(A=4 \times 5 \sqrt{1-\left(\frac{k}{4}\right)^{2}} \cdot \mathrm{k}\)
Now, differentiating w.r.t. k, we get -
\(\frac{\mathrm{dA}}{\mathrm{dk}}=20 \cdot \mathrm{k} \cdot\left[\frac{1}{2}\left[1-\left(\frac{\mathrm{k}}{4}\right)^{2}\right]^{-1 / 2}\left[0-\frac{2 \mathrm{k}}{16}\right]\right]+\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(=(20)\left[-\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}}+\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\right]\)
For max. k,
\(\frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}}=\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(\Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=1-\left(\frac{\mathrm{k}}{4}\right)^{2}\)
\(\Rightarrow 2\left(\frac{\mathrm{k}}{4}\right)^{2}=1 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=\frac{1}{2}\)
\(\therefore \mathrm{k}=\frac{4}{\sqrt{2}}=2 \sqrt{2}\)
So, we get
\(\frac{\mathrm{h}^{2}}{25}=1-\frac{\mathrm{k}^{2}}{16}=1-\frac{8}{16}=\frac{1}{2}\)
\(\therefore \quad \mathrm{h}^{2}=\frac{25}{2}\)
Or \(\quad \mathrm{h}=\frac{5}{\sqrt{2}}\)
Dimensions of the rectangle are
\(\begin{array}{ll}\therefore \quad 2 \mathrm{k} =2 \times 2 \sqrt{2}=4 \sqrt{2} \\ \quad 2 \mathrm{~h} =5 \sqrt{2}\end{array}\)
\(2 h=5 \sqrt{2}\)
So, the dimensions are \((4 \sqrt{2}, 5 \sqrt{2})\) units.

MHT CET-2021

Application of Derivatives

85739 A wire of length 20 units is divided into two parts such that the product of one part and cube of the other part is maximum, then product of these parts is

1 5 units

2 70 units

3 75 units

4 15 units

MHT CET-2021

Application of Derivatives

85740 Let \(f(x)=x^{2} \log x, x>0\). Then the minimum value of \(f\) is

1 \(\frac{1}{\sqrt{\mathrm{e}}}\)

2 \(2 \mathrm{e}\)

3 \(-2 \mathrm{e}\)

4 \(\sqrt{\mathrm{e}}\)

5 \(\frac{-1}{2 \mathrm{e}}\)

Kerala CEE-2021

Application of Derivatives

85741 The length of the longest size rectangle of maximum area that can be inscribed in a semicircle of radius 1 , so that 2 vertices lie on the diameter, is:

1 \(\sqrt{2}\)

2 2

3 \(\sqrt{3}\)

4 \(\frac{\sqrt{2}}{3}\)

5 \(\frac{-2}{\sqrt{3}}\)

Kerala CEE-2004

Application of Derivatives

85737 Let \(A D\) and \(B C\) be two vertical poles at \(A\) and \(B\) respectively on a horizontal ground, If \(A D=8 \mathrm{~m}\), \(B C=11 \mathrm{~m}\) and \(A B=10 \mathrm{~m}\), then the distance (in meters) of point \(M\) and \(A B\) from the point \(A\) such that \(\mathrm{MD}^{2}+\mathrm{MC}^{2}\) is minimum is

1 8

2 5

3 4

4 7

MHT CET-2022

Application of Derivatives

85738 A rectangle of maximum area is inscribed in an ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\), then its dimensions are

1 \(4 \sqrt{2}, 5 \sqrt{2}\) (units)

2 \(\sqrt{2}, 5 \sqrt{2}\) (units)

3 \(4 \sqrt{2}, \sqrt{2}\) (units)

4 \(4 \sqrt{2}, 6 \sqrt{2}\) (units)

Explanation:

(A) : Given the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Let, there is a point \((\mathrm{h}, \mathrm{k})\) on the ellipse. So, we get \(\frac{\mathrm{h}^{2}}{25}+\frac{\mathrm{k}^{2}}{16}=1\)
\(\therefore \frac{\mathrm{h}^{2}}{25}=1-\left(\frac{\mathrm{k}}{4}\right)^{2} \Rightarrow \mathrm{h}=5 \sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\)

Area of the rectangle A \(=2 \mathrm{~h} \times 2 \mathrm{k}\)
\(=4 \mathrm{hk} \text {. }\)
Putting the value of \(h\). we get -
\(A=4 \times 5 \sqrt{1-\left(\frac{k}{4}\right)^{2}} \cdot \mathrm{k}\)
Now, differentiating w.r.t. k, we get -
\(\frac{\mathrm{dA}}{\mathrm{dk}}=20 \cdot \mathrm{k} \cdot\left[\frac{1}{2}\left[1-\left(\frac{\mathrm{k}}{4}\right)^{2}\right]^{-1 / 2}\left[0-\frac{2 \mathrm{k}}{16}\right]\right]+\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(=(20)\left[-\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}}+\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\right]\)
For max. k,
\(\frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}}=\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(\Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=1-\left(\frac{\mathrm{k}}{4}\right)^{2}\)
\(\Rightarrow 2\left(\frac{\mathrm{k}}{4}\right)^{2}=1 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=\frac{1}{2}\)
\(\therefore \mathrm{k}=\frac{4}{\sqrt{2}}=2 \sqrt{2}\)
So, we get
\(\frac{\mathrm{h}^{2}}{25}=1-\frac{\mathrm{k}^{2}}{16}=1-\frac{8}{16}=\frac{1}{2}\)
\(\therefore \quad \mathrm{h}^{2}=\frac{25}{2}\)
Or \(\quad \mathrm{h}=\frac{5}{\sqrt{2}}\)
Dimensions of the rectangle are
\(\begin{array}{ll}\therefore \quad 2 \mathrm{k} =2 \times 2 \sqrt{2}=4 \sqrt{2} \\ \quad 2 \mathrm{~h} =5 \sqrt{2}\end{array}\)
\(2 h=5 \sqrt{2}\)
So, the dimensions are \((4 \sqrt{2}, 5 \sqrt{2})\) units.

MHT CET-2021

Application of Derivatives

85739 A wire of length 20 units is divided into two parts such that the product of one part and cube of the other part is maximum, then product of these parts is

1 5 units

2 70 units

3 75 units

4 15 units

MHT CET-2021

Application of Derivatives

85740 Let \(f(x)=x^{2} \log x, x>0\). Then the minimum value of \(f\) is

1 \(\frac{1}{\sqrt{\mathrm{e}}}\)

2 \(2 \mathrm{e}\)

3 \(-2 \mathrm{e}\)

4 \(\sqrt{\mathrm{e}}\)

5 \(\frac{-1}{2 \mathrm{e}}\)

Kerala CEE-2021

Application of Derivatives

85741 The length of the longest size rectangle of maximum area that can be inscribed in a semicircle of radius 1 , so that 2 vertices lie on the diameter, is:

1 \(\sqrt{2}\)

2 2

3 \(\sqrt{3}\)

4 \(\frac{\sqrt{2}}{3}\)

5 \(\frac{-2}{\sqrt{3}}\)

Kerala CEE-2004

Application of Derivatives

85737 Let \(A D\) and \(B C\) be two vertical poles at \(A\) and \(B\) respectively on a horizontal ground, If \(A D=8 \mathrm{~m}\), \(B C=11 \mathrm{~m}\) and \(A B=10 \mathrm{~m}\), then the distance (in meters) of point \(M\) and \(A B\) from the point \(A\) such that \(\mathrm{MD}^{2}+\mathrm{MC}^{2}\) is minimum is

1 8

2 5

3 4

4 7

MHT CET-2022

Application of Derivatives

85738 A rectangle of maximum area is inscribed in an ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\), then its dimensions are

1 \(4 \sqrt{2}, 5 \sqrt{2}\) (units)

2 \(\sqrt{2}, 5 \sqrt{2}\) (units)

3 \(4 \sqrt{2}, \sqrt{2}\) (units)

4 \(4 \sqrt{2}, 6 \sqrt{2}\) (units)

Explanation:

(A) : Given the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Let, there is a point \((\mathrm{h}, \mathrm{k})\) on the ellipse. So, we get \(\frac{\mathrm{h}^{2}}{25}+\frac{\mathrm{k}^{2}}{16}=1\)
\(\therefore \frac{\mathrm{h}^{2}}{25}=1-\left(\frac{\mathrm{k}}{4}\right)^{2} \Rightarrow \mathrm{h}=5 \sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\)

Area of the rectangle A \(=2 \mathrm{~h} \times 2 \mathrm{k}\)
\(=4 \mathrm{hk} \text {. }\)
Putting the value of \(h\). we get -
\(A=4 \times 5 \sqrt{1-\left(\frac{k}{4}\right)^{2}} \cdot \mathrm{k}\)
Now, differentiating w.r.t. k, we get -
\(\frac{\mathrm{dA}}{\mathrm{dk}}=20 \cdot \mathrm{k} \cdot\left[\frac{1}{2}\left[1-\left(\frac{\mathrm{k}}{4}\right)^{2}\right]^{-1 / 2}\left[0-\frac{2 \mathrm{k}}{16}\right]\right]+\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(=(20)\left[-\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}}+\sqrt{1-\left(\frac{\mathrm{k}}{(4)}\right)^{2}}\right]\)
For max. k,
\(\frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2} \cdot \frac{1}{\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}}=\sqrt{1-\left(\frac{\mathrm{k}}{4}\right)^{2}}\)
\(\Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=1-\left(\frac{\mathrm{k}}{4}\right)^{2}\)
\(\Rightarrow 2\left(\frac{\mathrm{k}}{4}\right)^{2}=1 \Rightarrow\left(\frac{\mathrm{k}}{4}\right)^{2}=\frac{1}{2}\)
\(\therefore \mathrm{k}=\frac{4}{\sqrt{2}}=2 \sqrt{2}\)
So, we get
\(\frac{\mathrm{h}^{2}}{25}=1-\frac{\mathrm{k}^{2}}{16}=1-\frac{8}{16}=\frac{1}{2}\)
\(\therefore \quad \mathrm{h}^{2}=\frac{25}{2}\)
Or \(\quad \mathrm{h}=\frac{5}{\sqrt{2}}\)
Dimensions of the rectangle are
\(\begin{array}{ll}\therefore \quad 2 \mathrm{k} =2 \times 2 \sqrt{2}=4 \sqrt{2} \\ \quad 2 \mathrm{~h} =5 \sqrt{2}\end{array}\)
\(2 h=5 \sqrt{2}\)
So, the dimensions are \((4 \sqrt{2}, 5 \sqrt{2})\) units.

MHT CET-2021

Application of Derivatives

85739 A wire of length 20 units is divided into two parts such that the product of one part and cube of the other part is maximum, then product of these parts is

1 5 units

2 70 units

3 75 units

4 15 units

MHT CET-2021

Application of Derivatives

85740 Let \(f(x)=x^{2} \log x, x>0\). Then the minimum value of \(f\) is

1 \(\frac{1}{\sqrt{\mathrm{e}}}\)

2 \(2 \mathrm{e}\)

3 \(-2 \mathrm{e}\)

4 \(\sqrt{\mathrm{e}}\)

5 \(\frac{-1}{2 \mathrm{e}}\)

Kerala CEE-2021

Application of Derivatives

85741 The length of the longest size rectangle of maximum area that can be inscribed in a semicircle of radius 1 , so that 2 vertices lie on the diameter, is:

1 \(\sqrt{2}\)

2 2

3 \(\sqrt{3}\)

4 \(\frac{\sqrt{2}}{3}\)

5 \(\frac{-2}{\sqrt{3}}\)

Kerala CEE-2004