Question

The cross-section of a tunnel has the form
of a rectangle surmounted by a semi-circle.
The perimeter of this cross section is 18 meters.
For what radius of the semi-circle will
the cross-section have maximum area?

Answer 1

you need to take 18 the perimeter including the semi-circle or just the perimeter of the rectangle?

Answer 2

edit: is 18 the perimiter of the whole thing semi-circle included or the perimeter of the box?

Answer 3

its the perimeter of the cross section, but i don't really know how the image looks

Answer 4

This is what it looks:

Answer 5

Now you have two variables, r and h (see image).
Because we know the perimeter, we can eliminate h.
The perimeter is 2r + 2h + halfcircle.
Because the perimeter of a circle is 2pi*r, a halfcircle has perimeter pi*r.
So:
18 = 2r + 2h + pi*r, so 2h = 18 - 2r - pi*r, and

\(h=9-r -\frac{\pi}{2}r=9-r(1+\frac{\pi}{2})\).

Answer 6

Why all this trouble? Well you have to minimize the area of the cross-section.
This area is the area of a rectangle with dimensions 2r and h and a semicircle with radius r.
Area of rectangle is 2r*h.
Area of semicircle is pi*r²/2.

Total area: \(A(r, h)=2rh+\frac{\pi}{2}r^2\).

But: A is now a function of two variables. Because we now know that h depends on r, we don't need h.
So, let's substitute \(h=9-r(1+\frac{\pi}{2})\) in the formula for A:

\(A(r)=2r(9-r(1+\frac{\pi}{2}))+\frac{\pi}{2}r^2\).
You have to simplify this formula as much as possible.
When that is done, you can calculate A'(r) and solve A'(r)=0 to find the optimum.

Answer 7

wait so the last equation is the original one, and when i calculate its derivative, it will give me maximum area of the cross section?

Answer 8

Yes, after simplifying it, A'(r) = 0 gives the value of r with the largest cross-section.

Answer 9

what is the derivative suppose to give me?

Answer 10

\(A(r)=18r-2(1+\frac{\pi}{2})r^2+\frac{\pi}{2}r^2=18r-2r^2-\pi r^2+\frac{\pi}{2}r^2\).
\(A(r)=18r-2r^2-\frac{\pi}{2}r^2=18r-(2+\frac{\pi}{2})r^2\).

So it is a 2nd degree function.

Then \(A'(r)=18-2(2+\frac{\pi}{2})r=0\)
And: \(9=(2+\frac{\pi}{2})r\),

So: \(r=\dfrac{9}{2+\frac{\pi}{2}}=\dfrac{9}{\frac{4+\pi}{2}}=9 \cdot \dfrac{2}{4+\pi}\).

Now we found: \(r=\dfrac{18}{4+\pi}\).

This will be the value of r that gives the maximum cross-section, although we could make a sign-scheme of A' to see that this r gives a maximum indeed.

Answer 11

As you can see, such a short question generates a lot of work!

Answer 12

yeah its tons of work! wait so the derivative of the equation is 18r−(2+π2)r2??

Answer 13

Yes, if you mean π/2 by "π2".

Answer 14

yeah sorry thats what i meant!

Answer 15

and if i have to do the chart that has the value of -infinity and +infinity? like what is the range of the equation?

Answer 16

wait were did you get the 9= part?

Answer 17

By dividing by 2.

Answer 18

ouu okok! got it! wow thank you so much!

Answer 19

can you help me in another problem?

Answer 20

Is your "Best Response" button stuck? :D

Answer 21

Just kidding, I will help you anyway!

Answer 22

ahahaha! ummm the problem is:
Find the point on the graph of y=square root of x which is closest to the point (4,0)

Answer 23

So this is the situation: (see image)
You have to find the position of P, when AP is shortest.

Answer 24

okk and how do i do that?

Answer 25

You have to make a function that describes the length of AP.
If P=(a, √a) and A=(4, 0), how do you calculate the length of AP?

Answer 26

AP is the line right?

Answer 27

i think i could do it with a triangle but I'm not sure

Answer 28

See the image, AP is indeed the line you need.
And yes, a triangle (and the Pythagorean Theorem) gives you the distance.
Can you set up a formula?

Answer 29

i know it has to have 4^2 but i don't really know how to get the other numbers