What Are the Dimensions to Minimize Fence Length for a 384 ft² Study Area?

[jenc305]

A rectangular study area is to be enclosed by a fence and divided into two equal parts, with a fence running along the division parallel to one of the sides. If the total area is 384 ft^2, find the dimensions of the study area that will minimize the total length of the fence. How much fence will be required?

This is what I have so far:
P=3y+4x
y=384ft^2/x
P=3(384ft^2)/x+4x
P=1152ft^2/x+4x

I'm not sure what to do next.

Thanks for the help!

 

[Galileo]

How did you get P=3y+4x?
Don't you have P=3x+2y? (Twice the length, twice the width and one division equal to either x or y?)

You have to minimize P, so try differentiating it to find the extrema.

 

[wais]

To solve this problem, we need to use the derivative to find the minimum value of P. We know that the total area is 384 ft^2, so we can set up an equation: 384 = x*y, where x and y are the dimensions of the study area. We also know that the fence will divide the area into two equal parts, so we can set up another equation: y = x/2.

Substituting this into our first equation, we get 384 = x*(x/2), or 384 = x^2/2. Solving for x, we get x = 24 ft. Since y = x/2, y = 12 ft. Therefore, the dimensions of the study area that will minimize the total length of the fence are 24 ft by 12 ft.

To find the total length of the fence, we can use the perimeter formula: P = 2x + 3y. Substituting our values, we get P = 2(24 ft) + 3(12 ft) = 48 ft + 36 ft = 84 ft. So, 84 ft of fence will be required to enclose and divide the study area into two equal parts.

I hope this helps and clarifies the process for solving this problem using calculus. Keep practicing and you'll get the hang of it!