# Challenging Question For SMARTIES

Explain How you get the answer,
The following practive field diagram shows a rectangle with a semi circle at each end. (you will have to draw the diagram 4sided rectangle with 2 semi circles on each end)

The track and field coach wants two laps around the field to be 1000m. The physical education department needs a rectangle field that is as large as possible. Determine the dimensions of the track that will maximize the entire enclosed area. Do these dimensions meet the needs of the track coach and the physical education department? Explain?

GOOD LUCK! I TRIED but Its COMPLICATED :uhh:

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James R
Homework Helper
Gold Member
Take the length of the rectangular part to be L, and the radius of each semicircular end to be r.

The distance round the track is:

$$2L + 2 \pi r = 500$$

The area of the track is:

$$A = 2rL + \pi r^2$$

You want to maximize A, subject to the condition on the length of the track.

Can you do it from here?

James did the hard part.
to find the maximum (or minimum) you set
$$\frac{dA}{dr} = 0$$

from James's first equation we have that
$$2L = 500-2\pi r$$

substitute this into his second equation to get
$$A = 500r - \pi r^2$$

now find the derivative
$$\frac{dA}{dr} = 500 - 2\pi r$$ (*)
$$\frac{d^2A}{dr^2} = -2\pi <0$$
so we see it is a maximum
we set * equal to zero for the maximum
=>
$$500-2\pi r=0$$

so the dimensions of the track that will maximize the entire enclosed area are
$$r=\frac{500}{2\pi}$$

where
$$L=250-\pi r$$

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BobG
Homework Helper

The rectangular area needs to be as big as possible. If you carry vladimir69's last two steps all the way through, you'll notice the straightaways have a length of zero. In other words, to maximize the area enclosed by the track, the track should be circular.

That's not what the question asked. You could still fit a square field inside the circle (that's a whole new problem), but it would be smaller than the largest rectangle that could fit inside the track if the track had straightaways.

Modify the area equation to:

$$A=2rL$$

You still use James R's equation for finding the perimeter of the track. The key is to solve for one variable in terms of the other. Then you can substitute that solution into the equation for area.

A little more help plz

ok I have never heard of the word derivitive and to find the max area I think I should be using the completing the square method since this is talked about in the unit. With the new area=2rL and the distance around the track 2L+2pie r=500 how can I substitute one equation into the other? Then what do I do next?

Diane_
Homework Helper
Solve your second equation (the one equal to 500) for one of the variables - which one doesn't really matter. Substitute that into the first equation, and you'll have a formula for the area that is quadratic in whichever variable you didn't solve for.

For instance - say you solve the second equation for r. When you substitute that into the first equation, you'll have the area as a function of L, and L will be quadratic.

Do you see, then, that you'll have a parabola? From that, you should be able to determine the value of (in our example) L that gives you the maximum area. From there, it's just a hop, skip and a jump (well, a hop and skip, anyway) to what that area is and the value of r.

from the second equation does L=250-pie r? or do i have to keep this as L=(500-2pie r)/2 nothing makes sense I used the first equation L=250-pie r and then plugged it into the area equation and got A=2r(250-3.14) I expanded this and got a quadratic fuction. Using this quadratic function I completed the square and solved for r, but when I used the value of r=80 or r=-7.03 (i used the first one 80) to find what the L= it did not work L=0.1? why What is going on? I think what I did was to find the max area of the rectangle in the middle of the track, but after I do find this I dont even know how to find out if the track is really 500 m? Please help me someone I am so confused.

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BobG
Homework Helper
A= L*2r

So, when you plugged in (L = 250 - pi r) for L you should have gotten:

$$A=(250-\pi r)*2r$$
$$A= 500r -2 r^2$$

or:

$$0=-2r^2 + 500r - A$$

If you look at the first equation, it's even easier. You have two points where the Area will be zero: when r = 0 or when $$\pi r = 250$$

Knowing the properties of a parabola (it's symmetric), you know your vertex is midway between the two. Since the most significant term (-r^2) is negative, the parabola is concave down. That means the vertex is your max.

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This online tutor told me that the equation A=2rL is not correct and that it should be A=2rl+pi r^2
So i subbed in L=250-pi r into the area equation and got 2r(250-pi r)+pi r^2 I expanded the brackets and got 500r-2pi r^2+pi r^2 I further simplified this to A=-pi r^2+500r now using this equation i complete the square and got A=-pi(r+80)^2+19854 therefore the vertex is (-80,19854) The a value is negative therefore this quadratic function has a maximum. The maximum is 19854. Now I solve for the radius by setting the eqn equal to 0, This is where i got stuck both values for r are negative -0.50, and -159 ? Where did I go wrong? BobG