# Thread: High school calculus problem. Is the question dodgy or am I missing something?

1. ## High school calculus problem. Is the question dodgy or am I missing something?

Hi guys.

So I've all up to (c), but in (d) the question talks about how x can be between 0 to 11 although in the context x must be less than 10 since the length of the wire is 10. This does not make sense to me. Or am I missing something?

Thanks

2. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

It would be a typo, it should be [0, 10].

3. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

If it was indeed a typo, would there be a maximum? I don't think there is because x is in (0,10), since (d) says two square are formed.

4. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

If it's possible to have an area from a finite amount, then it must have a maximum and a minimum.

Maxima and minima can occur either at stationary points or endpoints.

5. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

Were you able to do parts a, b, and c? If so, what answers did you get?

6. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

a. I could show this

b. $\dfrac{dA}{dx}=\dfrac{1}{8}(2x-10)=\dfrac{1}{4}(x-5)$

c. $x= 5$ since $\dfrac{d^{2}A}{dx^{2}}(x=5)=-\dfrac{5}{4}<0.$

7. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

So what now with question d?
I don't think it is possible to find a maximum that fits the context, although, it actually does happen and x=0 and x=10.
What now?

8. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

substitue them

9. ## Re: High school calculus problem. Is the question dodgy or am I missing something?

Originally Posted by willy0625
a. I could show this

b. $\dfrac{dA}{dx}=\dfrac{1}{8}(2x-10)=\dfrac{1}{4}(x-5)$

c. $x= 5$ since $\dfrac{d^{2}A}{dx^{2}}(x=5)=-\dfrac{5}{4}<0.$