SOLUTION: We have these word problems to do and they are so difficult for me to complete. Once I find the equations I can pretty much finish the problem but finding the equations is proving
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Question 248944: We have these word problems to do and they are so difficult for me to complete. Once I find the equations I can pretty much finish the problem but finding the equations is proving to be challenging. Please help.
Word Problem: A rancher can affors 300 feet of fencing to build a corral that's divided into two equal rectangles. What dimensions will maximize the corral's area?
I'm given a picture where the length is 2x and the width is y and the hint from my teacher was to use the distance formula. I'm completely lost. Thank you in advance for your help. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Word Problem: A rancher can afford 300 feet of fencing to build a corral that's divided into two equal rectangles.
What dimensions will maximize the corral's area?
I'm given a picture where the length is 2x and the width is y
:
I have no idea why you would use distance formula, perhaps he meant the perimeter formula
:
The fence equation
2(2x) + 3y = 300
4x + 3y = 300
3y = (300 - 4x)
y = 100 - x
:
Area = L * W
A = 2x * y
Replace y with(100-x)
A = 2x(100-x)
A = 200x - 2x(x)
A = 200x -x^2
:
A quadratic equation, the max area occurs at the axis of symmetry
The formula for that: x = -b/(2a)
In this equation a=-8/3; b=200
x =
x =
x = -200 *
x = +37.5
:
Find y:
y = 100 - x
y = 100 - *37.5
y = 100 - 50
y = 50 ft
;
Since the dimensions are are 2x by y, max area occurs when:
2(37.5) = 75 ft is the length and 50 ft is the width
:
Check the perimeter (fence length)
2(75) + 3(50) =
150 + 150 = 300
;
;
C
