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isyllus

Answer:

Maximum area of rectangle = 50 sq unit

Step-by-step explanation:

A rectangle inscribed in a semi-circle of radius 5.

Please find attachment for figure.

Let length of rectangle be 2y and width be x .

In ΔOAB, ∠OBA = 90°

[tex]OB^2+AB^2=OA^2[/tex]     (Using pythagoreous theorem)

[tex]y^2+x^2=5^2[/tex]

[tex]y^2=25-x^2[/tex]

[tex]y=\sqrt{25-x^2}[/tex]

Area of rectangle = Length x Width

                         A = 2y × x

                       [tex]A=2x\sqrt{25-x^2}[/tex]

Derivative above function w.r.t x

[tex]\dfrac{dA}{dx}=2\sqrt{25-x^2}-\dfrac{2x^2}{\sqrt{25-x^2}}[/tex]

For maxima/minima [tex]\dfrac{dA}{dx}=0[/tex]

[tex]2\sqrt{25-x^2}-\dfrac{2x^2}{\sqrt{25-x^2}}=0[/tex]

[tex]2x^2=25[/tex]

[tex]x=\dfrac{5}{\sqrt{2}}[/tex]

To check maximum/minima using double derivative test

[tex]\dfrac{d^2A}{dx^2}=\frac{2x\left(2x^{2}-75\right)}{\left(25-x^{2}\right)^{\frac{3}{2}}}[/tex]

At [tex]x=\dfrac{5}{\sqrt{2}}[/tex]  , [tex]\dfrac{d^2A}{dx^2}=-8<0[/tex]

Hence, the area is maximum when width, [tex]x=\dfrac{5}{\sqrt{2}}[/tex] and [tex]y=5\sqrt{2}[/tex]

Maximum area of rectangle, A= 2xy

[tex]A_{max} = 2\cdot \dfrac{5}{\sqrt{2}}\cdot 5\sqrt{2}[/tex]

[tex]A_{max}=50[/tex]

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aksnkj

Answer:

The area of the largest rectangle is 25 square units.

Step-by-step explanation:

It is given that the radius of the semicircle is [tex]r=5\texttt { unit}[/tex].

It is required to inscribe a rectangle of maximum area.

Consider a rectangle of sides [tex]x[/tex] and [tex]y[/tex] as shown in the given figure.

From the attached figure, [tex]AB=x[/tex] , [tex]OA=r[/tex] and [tex]OB=\dfrac {y}{2}[/tex].

Now, in the triangle OAB, apply Pythagoras theorem as,

[tex]OA^2=OB^2+AB^2\\\left (r^2 \right )=\left (\dfrac{y}{2} \right )^2+x^2\\25=\dfrac{y^2}{4}+x^2\\y=2\sqrt{25-x^2}[/tex]

So, the area of the rectangle will be,

[tex]A=xy\\A=x(2\sqrt{25-x^2})\\A=2x\sqrt{25-x^2}[/tex]

Differentiate the area with respect to [tex]x[/tex],

[tex]A=2x\sqrt{25-x^2}\\A'=2\sqrt{25-x^2}+2x(\dfrac{1}{2\sqrt{25-x^2}})(-2x)=0\\2\sqrt{25-x^2}=2x^2(\dfrac{1}{\sqrt{25-x^2}})\\25-x^2=x^2\\2x^2=25\\x=\dfrac{5}{\sqrt2}[/tex]

Differentiate the area twice as,

[tex]A'=2\sqrt{25-x^2}-(\dfrac{2x^2}{\sqrt{25-x^2}})\\A''=\dfrac{1}{\sqrt{25-x^2}}-4x\dfrac{1}{\sqrt{25-x^2}}+4x^3\dfrac{1}{({25-x^2)}^{3/2}}\\A''_{\left (x=\dfrac{5}{\sqrt2}\right )}=\dfrac{\sqrt2}{5}-4+\dfrac{\sqrt2}{5}\\A''_{\left (x=\dfrac{5}{\sqrt2}\right )}<0[/tex]

So, the double derivative of area function is negative and hence, the area will be maximum at [tex]x=\dfrac{5}{\sqrt2}\right[/tex].

Therefore, the maximum area of rectangle will be,

[tex]A=2x\sqrt{25-x^2}\\A=2\times \dfrac{5}{\sqrt2}\sqrt{25-\dfrac{25}{2}}\\A=25[/tex]

Therefore, the area of the largest rectangle is 25 square units.

For more details, refer the link:

https://brainly.com/question/14316282?referrer=searchResults

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