Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

minimizing area: A 36-in. piece of wire is cut into two pieces. One Piece is used to form a circle while the other is used to form a square. How should the wire be cut so that the sum of the areas is minimal? What is the minimum value?

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

|dw:1323260051758:dw| The sum of the areas is\[A(r)=\frac{1}{2}(2\pi)r^2 + (36-2\pi r)^2\]You want this to be a \(\textit{minimum}\). This will happen when the first derivative of A = 0. (though this could also be a maximum. To figure out which, you'd need the second derivative I think).
what do i do next?
That should get you a value for r.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Then you work out the length \(2\pi r\), and the length \((36 - 2\pi r)\)
Actually, according to my sanity test, that gives you lengths longer than 36, so it can't be right.
Nope, calculation error. Never mind.
i never did one of these problems and im sure he's going to put it on the test so i have to know how to do it -..-
i still dont know where to begin...
The total length of the wire is 36 in, right? That length will then turn into two lengths when you cut it in half. One length will become the circumference of your circle (\(2\pi r)\), and the other will become the perimeter of your square (36 - \(2\pi r\)). So now you have: |dw:1323260762343:dw| I actually made a mistake earlier, I forgot the perimeter of the square and the length of its side are not the same. So the area of the circle is \(\pi r^2\), and the area of the square is the side length squared. The side length is one quarter of the perimeter, so the area of the square is \(\left(\frac{1}{4}(36-2\pi r)\right)^2\) Thus, the sum of the two areas is....?
Precisely. So \[A(r)=\pi r^2 + 81-9\pi r+\frac{1}{4}\pi^2r^2\]when you simplify, and expand that bracket. What we have is a function, which defines the area in terms of the radius. To give you an idea of what that looks like, go to^2%2B%289-%5Cfrac{%5Cpi}{2}r%29^2 We're looking for that lowest point on the graph. Where is that point? Well it's at the point where the slope of the graph is zero. The slope of the graph is given by \[\frac{dA}{dr}\], and we want it to be equal to zero, so we say \[\frac{dA}{dr}=0\]
where did dA/dr come from?
Have you started calculus yet?
Good grief. I have no idea how to solve this problem without calculus. You've not seen anything like dy/dx yet?
no... my class is intermediate algebra
Well my apologies, I don't see how it's possible to solve this with just algebra.
hmm... can you type how to solve this your way and i can just write it down?
It's pretty advanced mathematics compared to what you're currently doing, there's just no way you'd be expected to do it this way.

Not the answer you are looking for?

Search for more explanations.

Ask your own question