A piece of wire of length 56 is​ cut, and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to ​(a) minimize and ​(b) maximize the combined area of the circle and the​ square?

Respuesta :

Answer:

1) Cut the wire at a distance of 31.36 and draw this length into a square and the remaining into a circle to minimize the area.

2) To maximize the area do not cut the wire but make the whole wire into a circle.

Step-by-step explanation:

let the wire be cut at a distance of x

We make a square of this wire

For the remaining length of (56 - x) we make a circle

Thus

Area of square = [tex]Area_{square}(side)^{2}\\\\=(x/4)^{2}[/tex]

Similarly the area of the circle equals

[tex]A_{circle}=\frac{(56-x)^{2}}{4\pi }[/tex]

Thus summing the areas we get

[tex]A=\frac{x^{2}}{16}+\frac{(56-x)^{2}}{4\pi }[/tex]

1) to find the critical points we differentiate the given area with respect to 'x'

Thus we have

[tex]\frac{dA}{dx}=\frac{d(\frac{x^{2}}{16}+\frac{(56-x)^{2}}{4\pi })}{dx}\\\\0=\frac{x}{8}-\frac{(56-x)}{2\pi }\\\\\frac{x}{8}+\frac{x}{2\pi }=\frac{28}{\pi }\\\\\therefore x=31.36[/tex]

Thus the length of the square wire should be x = 31.36

The length of the circular portion should be 56 - 31.36 = 24.64.

These lengths shall give the minimum combined area of the 2 figures.

2) Since the given function is a quadratic thus with the graph attached below we can see the maximum area occurs if all the wire is to be made into a circle thus to maximize the area the wire shall not be cut and wholly shaped into a circle.

Ver imagen A1peakenbe
ACCESS MORE
EDU ACCESS
Universidad de Mexico