Area of Rectangle As Function of x

[nycmathguy]

See attachment.

I say the area in terms of x and y is
A = length • width or x•y.

I know that y = sqrt{36 - x^2}.

A(x) = x•y

Replace y with sqrt{36 - x^2}.

A(x) = x•sqrt{36 - x^2}

Is this function?

If not, why not?

See attachment for domain.

I say the domain is (-00, -6] U [6, 00).

Yes?

 

[MathLover1]

A(x) = x•sqrt{36 - x^2}=> correct


=> x=6 or x=-6=> points are (6,0) and (-6,0)-> so, both are in domain
so, the domain is is [-6,6]
or
{x element R : -6<=x<=6}

 

[nycmathguy]

A(x) = x•sqrt{36 - x^2}=> correct


=> x=6 or x=-6=> points are (6,0) and (-6,0)-> so, both are in domain
so, the domain is is [-6,6]
or
{x element R : -6<=x<=6}

The domain is on the line y = 0 from -6 to 6.

 

[MathLover1]

The domain is on the line y = 0 from -6 to 6.

The domain is on the line x from -6 to 6 (take a look at graph)

 

[nycmathguy]

The domain is on the line x from -6 to 6 (take a look at graph)

The x-axis is the line y = 0.

Yes?

 

[MathLover1]

The domain is on the line y = 0 from -6 to 6.

correction to the formula for area, there was a mistake

length= x units left fro origin and x units right from origin: so L=x+x=2x

width=W=y=sqrt(36-x)^2

we know that area of rectangle is length*width

A=L*W

now, we can plug values

A=2x*sqrt(36-x)^2 see attached to understand why is 2x in formula

 

[nycmathguy]

correction to the formula for area, there was a mistake

length= x units left fro origin and x units right from origin: so L=x+x=2x

width=W=y=sqrt(36-x)^2

we know that area of rectangle is length*width

A=L*W

now, we can plug values

A=2x*sqrt(36-x)^2 see attached to understand why is 2x in formula

My equation is missing number 2. Your graph makes clear why 2x and not x alone is multiplied by
sqrt{36 - x^2} to form the needed formula.

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Note 2:

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