Calculus 1, finding maximum dimensions?

Question

I think I figured out the following Q by checking what anwser would work but I hope someone can explain it to me in steps.

A farmer has 160 feet of fencing to enclose 2 adjacent rectangular pig pens. What dimensions should be used so that the enclosed area will be a maximum?

After drawing the picture and naming the short sides x and the long sides y, I think
4x+3y=160
y=(160-4x)/3

I put y back in the first expression: 4x+3((160-4x)/3)=160

I need to do something with the derivative but what?

Out of the answers I can choose only 20ft by 80/3 ft gives me the total of 160ft.
How do I work this problem??

Answer 1

We are assuming that the pig pens are the same size in this problem.

Let's do it this way. You know that 4x+3y=160, so y = (160-4x)/3. You also know that what you are trying to do is maximize the area, A = 2xy = 2x(160-4x)/3 = 1/3(320x - 8x^2). This is the function you are seeking to maximize, so take the derivative of this function. f'(x) = 320/3 - 16/3 x. Set this equal to 0 and solve, finding x = 20. Now go back to the 4x + 3y = 160 to find y = 80/3.

Answer 2

The amount of fencing available is 160 feet, and this is rectangular pen which will be divided to form two pig pens.

Let the perimeter, P = 2y + 3x , P = 160

2y + 3x = 160 -----------------------(1)
y = (160 - 3x)/2
y = 80 - 1.5x -----------------(2)

The area of this pen, A = xy
= x (80 - 1.5x)
Thus the area, A = 80x - 1.5x^2

Since this area should be a maximum, we need to differentiate it and equate the result to zero.

dA/dx = 80 - 3x

80 - 3x = 0
3x = 80
x = 80/3 = 26, 6667 feet (put this resultin (2))

and y = 80 - 1.5x
y = 80 - 1.5(26.6667)
= 80 - 40
= 40 feet

I am very good with my math. You can rely on my answers ~ I just love calculus.

You can check these values( y = 40 and x = 26.6667) by substituting them in equation (1) to obtain 160.

Thus, 2y + 3x = 160
2(40) + 3(26.6667) = 160 ----------------Note: 26.667 = 80/3

Answer 3

in case you chop back squares of section "x" from each and each nook, the backside of the appropriate shape would be 3-2x and 7-2x on an identical time as the top would be x, so the quantity = x*(3-2x)*(7-2x) = 4x^3 - 20x^2 + 21x Differentiate wrt x = 12x^2 - 40x + 21, remedy for 0 will supply 12x^2 - 40x + 21=0 x=(40+sqrt(592))/24 (A) or x = (40-sqrt(592))/24 (B) 2nd differential of 4x^3 - 20x^2 + 21x = 24x - 40 At (A) the 2nd differential evaluates to sqrt(592) > 0 and at (B) it evaluate to -sqrt(592) consequently x=(B) = 0.653 is a maxima === optimum volume and the aspects of container are length = 7-2x = 5.694 width = 3-3x = a million.694 top = x = 0.653 volume = 6.298