Optimization Math Help: Finding Minimum Enclosed Area with Wire Cut

[fghtffyrdmns]

You could use the quotient rule if you don't mind working harder than you need to. For example, you could use the quotient rule to differentiate f(x) = x/4, but that would be silly. Instead, you could notice that x/4 = (1/4)x, so d/dx(x/4) = d/dx(1/4*x) = 1/4 * d/dx(x), using the constant-multiple rule.

In your case d/dx[(8 - x)²/16 + x²/(4pi)] = (1/16) d/dx(8 - x)²) + 1/(4pi)*d/dx(x²). (I also used the sum rule for derivatives.)

Yes, that is what I mean :)

I got x = --
Edit: wrong answer.

 

[Mark44]

We have A(x) = (1/16)(8 - x)² + (1/(4pi)x².
A'(x) = ?

Your answer should be an equation.

 

[fghtffyrdmns]

We have A(x) = (1/16)(8 - x)² + (1/(4pi)x².
A'(x) = ?

Your answer should be an equation.

A'(x) = $$\frac{-16+2x}{16} + \frac{x}{2\pi}$$

However, I have to set the first derivative to 0 so I can find at which x value it should be cut so the area is small as possible (critical point - max/min value).

 

[Mark44]

Your A'(x) is correct, but it would be better to simplify it. You can rewrite it as A'(x) = (x - 8)/8 + x/(2pi).

There is also a domain to consider, since A(x) is not defined for all real numbers. For example, even though you can calculate a value for A(-8), you can't cut the wire at -8 cm.

Your critical point will come at a point where A'(x) = 0 or at an endpoint of the domain. What is the implied domain for A(x) = (1/16)(8 - x)² + (1/(4pi)x²?

 

[fghtffyrdmns]

0<x<8 (cannot be equal to since it makes no sense to cut 8 cm or 0 cm.) X is the element of any real number is the domain.

So I can't solve now?

Edit: 0 and 8 do not work .

 

[Mark44]

You can choose to not make the cut at all, so that you make only a square or only a circle. So the domain of A(x) is {x in the real numbers| 0 <= x <= 8}.

Now finish the problem. Be sure to check x = 0 and x = 8 after you have solved A'(x) = 0. That way you can be sure you have found the absolute maximum area.