a farmer has 160 ft of fence and wants a pen to adjoin to the whole side of the 130 ft barn. what should the dimensions be for maximum area?
If the pen has to be joined to the whole side of the barn, there goes 130 ft. He/she has only 30 ft left for the two sides. The dimensions are therefore 15 ft * 130 ft = 1950 sq.ft.
There is no calculations involving calculus, unless I have misinterpreted the question.
A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $30/ft and on the other three sides by a metal fence costing $20/ft. If the area of the garden is 108 square feet, find the dimensions of the garden that minimize the cost.
A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $30/ft and on the other three sides by a metal fence costing $20/ft. If the area of the garden is 108 square feet, find the dimensions of the garden that minimize the cost.
Length of side with bricks x
Length of adjacent side y
To find the dimensions that will maximize the area, we need to consider that the pen will be attached to one side of the barn. Let's assume that the length of the barn is the side where the pen will be attached.
Let's denote the width of the pen as 'x' and the length of the barn as 'y'. We can write the equation for the perimeter of the pen as:
Perimeter = 2x + y
Since the farmer has 160 ft of fence, the total perimeter of the pen (including the attached side to the barn) is 160 ft. Therefore, we can write:
2x + y = 160
We also know that the barn has a length of 130 ft. So, we have another equation:
y = 130
To maximize the area, we need to find the dimensions that satisfy both equations.
We can solve the first equation for 'y' in terms of 'x':
y = 160 - 2x
Now substitute this value into the second equation:
160 - 2x = 130
Simplify the equation:
2x = 160 - 130
2x = 30
x = 30/2
x = 15
Now substitute the value of 'x' back into the equation 'y = 160 - 2x':
y = 160 - 2(15)
y = 160 - 30
y = 130
Therefore, the dimensions of the pen that will maximize the area are 15 ft (width) and 130 ft (length).