A rectangle is constructed with its base on the diameter of a semicircle with radius 23 and with its two other vertices on the semicircle. What are the dimensions of the rectangle with maximum area?

Given : A rectangle is constructed with its base on the diameter of a semicircle with radius 23 and with its two other vertices on the semicircle.

To find : What are the dimensions of the rectangle with maximum area?

Solution :

Let origin be the center of the circle with radius 23.

The equation of the semicircle is [tex]x^2+y^2=23^2[/tex]

[tex]x^2+y^2=529[/tex]

Let (z,0) and (-z,0) be the points on the diameter of the semicircle.

∴ [tex]z^2+y^2=529[/tex]

[tex]y^2=529-z^2[/tex]

[tex]y=\pm\sqrt{529-z^2}[/tex]

Since the semicircle in on y-axis so coordinates of point on the semicircle are

[tex](z,\sqrt{529-z^2})(-z,\sqrt{529-z^2})[/tex]

So, The length of the rectangle is [tex]L=2z[/tex]

Breadth of the rectangle is [tex]B=\sqrt{529-z^2}[/tex]

Area of the rectangle is [tex]A=L\times B[/tex]

[tex]A=2z\times\sqrt{529-z^2} [/tex]

To maximum we have to derivate w.r.t z to find critical point.

[tex]\frac{dA}{dz}=2(\sqrt{529-z^2})+2z\times \frac{1}{2}\times (-2z)\times(\frac{1}{\sqrt{529-z^2}})[/tex]

[tex]0=2(\sqrt{529-z^2})-\frac{2z^2}{\sqrt{529-z^2}}[/tex]

[tex]\frac{2(529-z^2)-2z^2}{\sqrt{529-z^2}}=0[/tex]

[tex]1058-4z^2=0[/tex]

[tex]4z^2=1058[/tex]

[tex]z^2=\frac{529}{2}[/tex]

[tex]z=\sqrt{\frac{529}{2}}[/tex]

[tex]z=\frac{23}{\sqrt{2}}[/tex]

Substitute the value of z in the area and dimensions.

Length is [tex]L=2z[/tex]

[tex]L=2\times \frac{23}{\sqrt{2}}[/tex]

[tex]L=23\sqrt{2}[/tex]

Breadth is [tex]B=\sqrt{529-z^2}[/tex]

[tex]B=\sqrt{529-\frac{529}{2}}[/tex]

[tex]B=\sqrt{\frac{1058-529}{2}}[/tex]

[tex]B=\sqrt{\frac{529}{2}}[/tex]

[tex]B=\frac{23}{\sqrt{2}}[/tex]

Area is [tex]A=23\sqrt{2}\times \frac{23}{\sqrt{2}}[/tex]

[tex]A=23\times23[/tex]

[tex]A=529[/tex]

Therefore, The area of the rectangle is 529 unit square.

MrRoyal MrRoyal · Answer 2 · 2021-11-04T09:35:40+01:00

The maximum area of a shape is the largest area of the shape.

The dimensions of the rectangle are [tex]\mathbf{L = 23\sqrt 2}[/tex] and [tex]\mathbf{W = \frac{23\sqrt2}{2}}[/tex]
The maximum area of the rectangle is 529 square units.

The given parameter is:

[tex]\mathbf{r = 23}[/tex]

The equation of a semicircle that passes through the origin is:

[tex]\mathbf{x^2 + y^2 = r^2}[/tex]

Substitute 23 for r

[tex]\mathbf{x^2 + y^2 = 23^2}[/tex]

[tex]\mathbf{x^2 + y^2 =529}[/tex]

From the question, we understand that the other two vertices are on the semicircle.

So, the points on the semicircle are: [tex]\mathbf{(\pm a, 0)}[/tex]

Where:

[tex]\mathbf{L =2a}[/tex] --- length of the rectangle

[tex]\mathbf{W = y}[/tex] --- width of the rectangle

So, we have:

[tex]\mathbf{a^2 + y^2 =529}[/tex]

Rewrite as:

[tex]\mathbf{y^2 =529 - a^2}[/tex]

Solve for y

[tex]\mathbf{y =\sqrt{529 - a^2}}[/tex]

So, we have:

[tex]\mathbf{(\pm a, 0) = (a, \sqrt{529 - a^2}),(-a, \sqrt{529 - a^2})}[/tex]

The area of the rectangle is calculated as:

[tex]\mathbf{A = LW}[/tex]

This gives

[tex]\mathbf{A = 2ay}[/tex]

Substitute [tex]\mathbf{y =\sqrt{529 - a^2}}[/tex]

[tex]\mathbf{A = 2a\sqrt{529 - a^2}}[/tex]

Rewrite as:

[tex]\mathbf{A = 2a(529 - a^2)^{\frac{1}{2}}}[/tex]

Differentiate using chain rule

[tex]\mathbf{A' = 2 \times (529 - a^2)^\frac{1}{2} + 2a \times \frac 12 \times (-2a) \times (529 - a^2)^{-\frac{1}{2}}}[/tex]

Set to 0

[tex]\mathbf{2 \times (529 - a^2)^\frac{1}{2} + 2a \times \frac 12 \times (-2a) \times (529 - a^2)^{-\frac{1}{2}} = 0}[/tex]

Divide through by 2

[tex]\mathbf{ (529 - a^2)^\frac{1}{2} + a \times \frac 12 \times (-2a) \times (529 - a^2)^{-\frac{1}{2}} = 0}[/tex]

Multiply both sides by [tex]\mathbf{ (529 - a^2)^{-\frac{1}{2}}}[/tex]

[tex]\mathbf{ 529 - a^2 + a \times \frac 12 \times (-2a) = 0}[/tex]

[tex]\mathbf{ 529 - a^2 - a^2 = 0}[/tex]

[tex]\mathbf{ 529 - 2a^2 = 0}[/tex]

Collect like terms

[tex]\mathbf{ 2a^2 = 529}[/tex]

Divide both sides by 2

[tex]\mathbf{ a^2 =\frac{ 529}{2}}[/tex]

Take square roots

[tex]\mathbf{a =\sqrt{\frac{ 529}{2}}}[/tex]

[tex]\mathbf{a =\frac{23}{\sqrt2}}}[/tex]

Rationalize

[tex]\mathbf{a =\frac{23\sqrt2}{2}}}[/tex]

Recall that:

[tex]\mathbf{L =2a}[/tex]

[tex]\mathbf{L = 2 \times \frac{23\sqrt 2}{2}}[/tex]

[tex]\mathbf{L = 23\sqrt 2}[/tex]

Recall that:

[tex]\mathbf{W = y}[/tex]

[tex]\mathbf{W = \sqrt{529 - a^2}}[/tex]

[tex]\mathbf{W = \sqrt{529 - (\frac{23\sqrt2}{2})^2}}[/tex]

[tex]\mathbf{W = \sqrt{529 - \frac{529}{2}}}[/tex]

[tex]\mathbf{W = \sqrt{\frac{529}{2}}}[/tex]

[tex]\mathbf{W = \frac{23}{\sqrt2}}[/tex]

Rationalize

[tex]\mathbf{W = \frac{23\sqrt2}{2}}[/tex]

Recall that: [tex]\mathbf{A = 2a\sqrt{529 - a^2}}[/tex].

So, we have:

[tex]\mathbf{A =2 \times \frac{23\sqrt2}{2} \times \sqrt{529 - (\frac{23\sqrt2}{2})^2}}[/tex]

[tex]\mathbf{A =2 \times \frac{23\sqrt2}{2} \times \sqrt{529 - \frac{529}{2}}}[/tex]

[tex]\mathbf{A =2 \times \frac{23\sqrt2}{2} \times \sqrt{\frac{529}{2}}}[/tex]

[tex]\mathbf{A =2 \times \frac{23\sqrt2}{2} \times \frac{23}{\sqrt2}}}[/tex]

[tex]\mathbf{A =1 \times \frac{23}{1} \times \frac{23}{1}}[/tex]

[tex]\mathbf{A =529}[/tex]

So, the area of the rectangle is 529 square units.

A rectangle is constructed with its base on the diameter of a semicircle with radius 23 and with its two other vertices on the semicircle. What are the dimensions of the rectangle with maximum​ area?

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A rectangle is constructed with its base on the diameter of a semicircle with radius 23 and with its two other vertices on the semicircle. What are the dimensions of the rectangle with maximum area?