Mathematics, 25.11.2019 21:31 jibjabsalar1212

Find the area of the largest rectangle that can be inscribed in a semicircle of radius 5.

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Answer from: UncleVictor5188

The area of the largest rectangle is 25 square units.

Step-by-step explanation:

It is given that the radius of the semicircle is $r=5\texttt { unit}$ .

It is required to inscribe a rectangle of maximum area.

Consider a rectangle of sides and as shown in the given figure.

From the attached figure, AB=x , OA=r and $OB=\dfrac {y}{2}$ .

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

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

So, the area of the rectangle will be,

$A=xy\\A=x(2\sqrt{25-x^2})\\A=2x\sqrt{25-x^2}$

Differentiate the area with respect to ,

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

Differentiate the area twice as,

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

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

Therefore, the maximum area of rectangle will be,

$A=2x\sqrt{25-x^2}\\A=2\times \dfrac{5}{\sqrt2}\sqrt{25-\dfrac{25}{2}}\\A=25$

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

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Find the area of the largest rectangle that can be inscribed in a semicircle of radius 5.

Answer from: ramireztony741

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°

OB^2+AB^2=OA^2 (Using pythagoreous theorem)

y^2+x^2=5^2

y^2=25-x^2

$y=\sqrt{25-x^2}$

Area of rectangle = Length x Width

A = 2y × x

$A=2x\sqrt{25-x^2}$

Derivative above function w.r.t x

$\dfrac{dA}{dx}=2\sqrt{25-x^2}-\dfrac{2x^2}{\sqrt{25-x^2}}$

For maxima/minima $\dfrac{dA}{dx}=0$

$2\sqrt{25-x^2}-\dfrac{2x^2}{\sqrt{25-x^2}}=0$

2x^2=25

$x=\dfrac{5}{\sqrt{2}}$

To check maximum/minima using double derivative test

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

At $x=\dfrac{5}{\sqrt{2}}$ , $\dfrac{d^2A}{dx^2}=-8$

Hence, the area is maximum when width, $x=\dfrac{5}{\sqrt{2}}$ and $y=5\sqrt{2}$

Maximum area of rectangle, A= 2xy

$A_{max} = 2\cdot \dfrac{5}{\sqrt{2}}\cdot 5\sqrt{2}$

$A_{max}=50$

Answer from: Quest

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

Answer from: Quest

If you need an equation, it would be y=2x+8, y would be how long it grew, x being weeks.if you need how long he was asleep, it would be 4 weeks if he started with no beard, the equation would be x=8÷2if he did have a beard before hand, the equation would be 2w+x=8

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Find the area of the largest rectangle that can be inscribed in a semicircle of radius 5....

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