Calculus optimization?

Question

With the given function, y=12-x^2, with y greater than or equal to 0, a rectangle is to be inscribed in a semicircle of radius, r. What are the dimensions of the rectangle if its area is to be maximized?

Answer 1

Note that your problem is a bit contradictory. If it's a parabola you want, see below. If it's a circle you want, the answer is a half-square (a full square, if you looked at the whole circle), due to a simple geometric argument. I've gone ahead and done out both arguments using calculus. I hope this helps.
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Write out the vertices of some rectangle:

(-x,0) (x,0)

(-x,12-x^2) (x, 12-x^2)

Now notice the length: 2x
and the height: 12-x^2

Multiply for the area: 24x - x^3

Take the derivative of area:
24 - 3 x^2

Solve when that's equal to 0:

24 - 3x^2 = 0

8 = x^2

x = sqrt(8) = 2sqrt(2)

And so the dimensions are:

2x = 4sqrt(2)
and
12 - (2sqrt(2) )^2 = 4

With area: 16sqrt(2)


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Addition: I decided I'd go ahead and do the case when it actually is a circle, assuming what you meant to write was sqrt(12-x^2).

Write out the vertices of some rectangle:

(-x,0) (x,0)

(-x,sqrt(12-x^2)) (x, sqrt(12-x^2))

Now notice the length: 2x
and the height: sqrt(12-x^2)

Multiply for the area: 2xsqrt(12-x^2)

Take the derivative of area:
4(6-x^2) / sqrt(12-x^2)

Solve when that's equal to 0:

4(6-x^2) = 0

x=sqrt(6)

And so the dimensions are:

2x = 2sqrt(6)
and
sqrt(12 - (sqrt(6) )^2) = sqrt(6)

With area: 12

Answer 2

Don't understand the significance of y=12-x^2 which is a parabola. The question is what are the dimensions of the largest rectangle that can be inscribed in a semicircle of radius r?

The answer is the width must be r*sqrt(2), and height must be r*sqrt(2)/2. This gives an area of r^2.

Answer 3

1) It is simple; ==> First decide the objective of the problem; here it is paper dimensions, so that its area will be minimum; then next comes under what constraints this has to be done? ==> Well, here On the selected paper, you are leaving some margins all over and arriving at the area for printing, which is given some fixed value. 2) Now, we are clear what we want; How to proceed for building mathematical equations, so that it can be solved for getting the end answer: 3) Here we are given the printing area as 50 sq in which is constant; hence let us start from this data; ==> Let the dimension of the printing area be x in (height wise) by y in (width wise) ==> Printing area = xy = 50; == y = 50/x -------(1) 4) There is a margin of 4 in each at top and bottom are provided; hence overall height of the paper is "x + 8" in; Similarly a margin of 2 in is provided on either sides; ==> Overall width = y + 4 in 5) Hence the area of the paper is = (x+8)(y+4) 6) Substituting for y from equation (1), A (x+8)(50/x + 4) = 50 + 400/x + 4x + 32 7) Hence the function to be minimized is: A = 82 + 4x + 400/x 8) Differentiating this, A' = 0 + 4 - 200/x^2 = 4 - 400/x^2 9) Equating A' = 0, x^2 = 100; ==> x = +/- 10 in 10) But a dimension cannot be negative, hence we consider only + 10; So x = 10 in and y = 5 in 11) However we need to verify,whether this is minimum or maximum; for which we will apply 2nd derivative test; So again differentiating, A'' = 800/x^3; at x = 10, A'' is > 0; hence it is minimum Thus we conclude for the printing the area to be 50 sq in, under the given constraints of margins, the outer size of the paper must be 18 in by 9 in; So outer area is = 162 sq in. Wish you are explained; have a nice time.

Answer 4

You want to maximize area, so come up with a formula for area: A = x*y

You know y = 12-x^2, so A = x * (12-x^2)

Find the derivative of A and set it equal to 0 to find the maximum A.

You should find that x = 2 whey A' = 0.

Plug that into y = 12-x^2 to find y

x = 2
y = 8

Answer 5

let the center of the circle be at point (0,0) and the Radius r
let the rectangle rest on the x axis
let x be a point on the X axis where we build the side of the rectangle
the hight of this side will be h where h^2+x^2=r^2
or H=(r^2-x^2)^.5
the rectangle streaches to the negative values of x so the base is (2x) and the hught is H=(r^2-x^2)^.5
The area is base*side=2x*(r^2-x^2)^.5
to find the maximum area that can be covered you need to find the value of x that causes the derivative of the area with respect to x to be zero
personally i hate differentiating variables with fractional powers, so lets elliminate it
since the square of a variable is monotonous increasing i will investigate the square of the area of the triangle.
A^2= (2x*(r^2-x^2)^.5)^2=4x^2*(r^2-x^2) hence we elliminated the fractional power
dA^2(x)/dx=8x*(r^2-x^2)-4x^2*2x=8xR^2-8x^3-8x^3
=8x(R^2-2*x^2)
for Maximum Value we need dA^2(x)/dx to be zero
so either 8x is zero at x=0 (this is the minimum value)
OR (R^2-2*x^2)=0
this would imply 2x^2=R^2 and x= (R^2/2)^.5=R*(0.5)^.5
the base of the rectangle is 2*R*0.5^.5
The side h is H=(R^2-x^2)^.5=(R^2-(R^2)/2)=R*0.5^0.5
The Area is 2*R*0.5^.5*R*0.5^.5

Answer 6

With parabola a rectangle is to be inscribed in a semicircle?

Can you explain what's the meaning of such sentence?

Answer 7

Please add detail as to how the semicircle is related to the inverted parabola. Otherwise, as all the answers above prove, we are all guessing at what the problem IS.........

Answer 8

Is that supposed to be sqrt(12 - x^2)?

Answer 9

i think you take the derivative so the sides are 2x. making it a square which is a rectangle, squares will always have more area.