Derivatives of Polynomials, Sum and Constant Factor Rules?

Question

A farmer’s barn is 60 feet long on one side. He has 150 feet of fence and wishes to build two adjacent rectangular pens of equal area along that side of his barn. What should be the arrangement and dimensions of the pen to maximize the sum of the areas of the two pens?

 

Answer 1

30 by 30 ft each pen

900 ft^2 for each pen

w=30 ft, 3 fences of 30 ft

L=60 ft, 1 fence of 60ft with the barn side on the other side

L+3w = 60+3(30) = 150 feet of fence

total area =Lw = 60(30)

= 1800 ft^2 =maximum possible Area

if L<60 then A<1800

if L=55, w = (150-55)/3 = 95/3 = 32 2/3

A= 55(95/3) = 1741 2/3 ft^2

if L=50, w = (150-50)/3 = 33 1/3

A = 50(100/3 = 5000/3 = 16

L+3w = 150

L=150-3w

A=wL= w(150-3w)=150w-3w^2

take the derivative of A with respect to w and set =0

A'(w) = 150 -6w = 0

w = 150/6= 25 feet wide, L=150-3(25) = 75 ft but the barn side is only 60 feet

use another 45 feet to extend the barn side, but that leaves no fence left for the 3 widths and A=0

so L had to be 60 feet

leaving 150-60=90 feet for 3 fences, each 30 feet long

Unless, maybe some fence were used to extend the barn side's length but an extension less than 45 feet,

the extension = L-60

then

150 =L + L-60 + 3w = 2L-60+3w

2L = 210-3w

w = (210-2L)/3 = 70-2L/3

L = (210-3w)/2= 105 -3w/2

extend the barn side with 10 feet, L=60+10=70 ft

w then = (150-80)/3 =70/3=23 1/3 ft wide

A= 70(70)/3 = 4900/3 = 1633 1/3 ft^2

extend the barn side 5 ft, L=65 ft, w= 80/3 = 26 2/3 ft

A = 65(80/3)= 5200/3 = 1733 1/3 ft^2

extend the barn side 1 ft, L=61, w = 88/3

A= 61(88/3) = 1789 1/3 ft^2 < 1800 when L=60, w=30

extend the barn 2 ft, L=62, w = 86/3

A = 62(86/3) = 1777 1/3 ft^2

extend the barn side 3 ft, L=63, w = 84/3

A = 63(84/3) = 1764 ft^2

extend the barn side 4 ft, L= 64, w = 82/3

A=Lw = w(105-3w/2) = 105w - 3w^2/2

A' = 105 -3w = 0

w = 105/3 =35 feet wide

L = 150-3(35) = 150-105 =45

A= Lw = 45(35) = 1575 ft^2

L=59, w = (150-59)/3 = 91/3

A= 59(91/3)=1789 2/3

all less areas than 1800 with 30 by 30 pens