Asked by Sakura
Ok, so the problem was about a person with 300 feet of fence to make a rectangle enclose. The person also wanted to use the fence to split the enclose to two parts with the fence paralel to two of the sides. Find the dimensions
So I am confuse on how to solve this.
So I am confuse on how to solve this.
Answers
Answered by
Reiny
I will assume you want the area to be a maximum and you want the dimensions for that .
Did you make a sketch?
Let the width be x and the length be y ft
so 3x + 2y = 300
y = (300-3x)/2
area = xy
= x(300-3x)/2
= 150x - (3/2)x^2
Are you taking Calculus?
then
d(area) = 150 - 3x = 0 for a max/min
3x = 150
x = 50 , then y = (300 - 150)/2 = 75
the width is 50 ft and the length = 75 ft
check:
3(50) + 2(75) = 300
area = 3750 for those dimenstions
if x = 49 , then y = 76.5
49(76.5) = 3748.5 < 3750
if x=51 , then y = 73.5
51(73.5) = 3748.5 < 3750
My answer is correct
Did you make a sketch?
Let the width be x and the length be y ft
so 3x + 2y = 300
y = (300-3x)/2
area = xy
= x(300-3x)/2
= 150x - (3/2)x^2
Are you taking Calculus?
then
d(area) = 150 - 3x = 0 for a max/min
3x = 150
x = 50 , then y = (300 - 150)/2 = 75
the width is 50 ft and the length = 75 ft
check:
3(50) + 2(75) = 300
area = 3750 for those dimenstions
if x = 49 , then y = 76.5
49(76.5) = 3748.5 < 3750
if x=51 , then y = 73.5
51(73.5) = 3748.5 < 3750
My answer is correct
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