# calculus 1 help?

your neighbor has a total of 400 square feet of sod to place turf in his backyard. he wants to construct a rectangular fenced area that contains the sod and a circular pool. the pool will be tangent to both of the long sides of the rectangular fence. he asks you to determine the minimum perimeter of the fence such that the area of turf inside the rectangular fence and outside the pool is 400 square feet. using calculus and showing all your work, solve this problem.

Relevance
• Let x be the long side of the rectangle, y the short side.

radius of circle is y/2, area is π(y/2)² = (π/4)y²

area of rectangle = xy

area of turf is A = xy – (π/4)y² = 400

perimeter = P = 2x+2y (this is what we want to minimize)

xy – (π/4)y² = 400

xy = (π/4)y² + 400

x = (π/4)y + 400/y

substitute

P = 2y + 2x

P = 2y + 2[(π/4)y + 400/y]

P = 2y + (π/2)y + 800/y

to minimize, differentiate and set equal to 0

P' = 2 +  (π/2) – 800/y² = 0

800/y² = 2 + (π/2)

800 = (2 + (π/2))y²

y² = 800 / (2 + (π/2))

y = √(800 / (2 + (π/2))) = √224.0397

y = 14.97 ft

x = (π/4)y + 400/y

x = (π/4)(14.97) + 400/(14.97)

x = 11.76 + 26.72 = 38.48 ft

P = 2x+2y = 2•38.48 + 2•14.97 = 106.89 ft

check

A rectangle = 14.97 x 38.48 = 576

A circle = 176

A turf = 576 – 176 = 400 • Login to reply the answers
• width of fence = diameter of pool

Sod area = Lw - πr^2

400 = 2Lr -πr^2

P = 2(L +w) = 2L +4r

(400 -πr^2)/2r = L

P = 2(400-πr^2)/2r +4r

P = 400/r -πr +4r

P' = -400/r^2 + (4-π)

r^2 = 400/(4-π) = 465.98

r = 21.6'

• Login to reply the answers
• If the circle is tangent to both of the long sides of the rectangle, then the short side is the diameter of the circle or the radius is half of this short side.

r = l/2

Due to how the fencing is arranged, there is no overlap or gaps.  We have the perimeter of the rectangle and the circumference of the circle added together.  So this amount of fencing in terms of three unknowns is:

f(r, l, w) = 2l + 2w + 2πr

We know the area of the rectangle is 400 ft², so we can create a third equation with this area:

A = lw

400 = lw

We now have three equations with four unknowns.  If we substitute what we know from the smaller two into the function we create a function with only one unknown:

Solving the last one for l in terms of w:

400 / w = l

Now we can substitute the (l/2) expression into the function for r:

f(r, l, w) = 2l + 2w + 2πr

f(l, w) = 2l + 2w + 2π(l/2)

f(l, w) = 2l + 2w + lπ

Now substitute (400/w) for l:

f(w) = 2(400/w) + 2w + (400/w)π

f(w) = 800/w + 2w + 400π/w

Add the two terms with the common denominator:

f(w) = 2w + 1200π/w

We now have a function with one unknown.  We can find w needed to make the minimum f() by solving for the zero of the first derivative:

f'(w) = 2 - 1200π/w²

0 = 2 - 1200π/w²

Let's move the term with "w" in it to the right side:

1200π/w² = 2

Multiply both sides by the denominator:

1200π = 2w²

Divide both sides by 2:

600π = w²

And get the square root of both sides:

w = ±√(600π)

w = ±√(100 * 6π)

w = ±10√(6π)

We can't have a negative width, so throwing that out:

w = 10√(6π)

That's the "w" that makes the minimum f().  We're only asked to solve for f(w) and no other variables, so:

f(10√(6π)) = 2(10√(6π)) + 1200π/(10√(6π))

Simplify:

f(10√(6π)) = 20√(6π) + 120π/√(6π)

Rationalizing the denominator:

f(10√(6π)) = 20√(6π) + 120π√(6π)/(6π)

The 6π's cancel out:

f(10√(6π)) = 20√(6π) + 20√(6π)

And this simplifies:

f(10√(6π)) = 40√(6π) ft

That's the minimum amount of fencing required.

• Login to reply the answers