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3 years ago

Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10.

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

5 0

Answer:

Step-by-step explanation:

That will be a square with a diagonal of 20

side length of 20sin45

and area of (20sin45)² = 200 units²

prove it you say?

Area of a rectangle is base times height

A = bh

With a radius of 10, the diagonals of any rectangle inscribed will be 20 units

20² = b² + h²

h = $\sqrt{400 - b^2}$

A = bh

A = b $\sqrt{400 - b^2}$

Area will be maximized when the derivative is set to zero

dA/db = $\sqrt{400 - b^2}$ - b²/ $\sqrt{400 - b^2}$

0 = $\sqrt{400 - b^2}$ - b²/ $\sqrt{400 - b^2}$

b²/ $\sqrt{400 - b^2}$ = $\sqrt{400 - b^2}$

b² = 400 - b²

2b² = 400

b² = 200

b = $\sqrt{200}$

h = $\sqrt{400 - b^2}$

h = $\sqrt{400 - \sqrt{200}^2 }$

h = $\sqrt{200}$

A = bh

A = $\sqrt{200}$ • $\sqrt{200}$

A = 200 units²

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Answer:

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Step-by-step explanation:

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3 years ago

Solved by substitution for <br>y=x-6<br>y=8x+1

GenaCL600 [577]

Here is your system of equations:

$\left \{ {{y=x-6} \atop {y=8x+1}} \right.$

When solving a system of equations, you can either eliminate or substitute, however, I will be substituting since that's what the question is asking.

In the first equation, if y is equal to x - 6, I can substitute it with the y in the second equation with 8x + 1.

Here is your new equation:

$x-6 = 8x+1$

Since you have variables on both sides, you have to cancel it on either side. In the equation on the left, the variable is x. To cancel it out, you have to do the opposite of it, which is -x, or negative x. That means you have to subtract x to both equations.

$x-6-x = 8x+1-x$

$-6=7x+1 \rightarrow 7x+1=-6$

Now you have one variable and two regular integers. You have to leave the variable x alone to find its value. Along with it is +1, or positive 1. To remove it, you have to do the opposite of it, which is -1 or subtract 1.

$7x+1-1=-6-1$