.
Instead of solving your problem, I'll give you the solution to a TWIN problem,
to give you an opportunity to learn the method and the subject.
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A farmer plans to fence a rectangular grazing area along a river with 300 yards of fence.
What is the largest area he can enclose?
Solution
Since one side is the river, the rectangle's fence perimeter will be
L + 2W = 300.
Hence, L = 300 - 2W.
Area = Length * Width.
Substitute (300-2W) for L:
A = W(300 - 2W)
A = -2W^2 + 300W.
It is a quadratic function. It has the maximum at x = -b/(2a), according to the general theory.
(See the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
in this site).
For our quadratic function the maximum is at
W = = = 75.
So, W = 75 yards is the width for max area.
Then the length is L = 300 - 2W = 300 - 2*75 = 150 yards.
Find the maximum area. It is
A = L*W = 150*75 = 11250 square yards.
The plot of the quadratic function for the area is shown below: y = area and x = width.
My other lessons in this site on finding the maximum/minimum of a quadratic function are
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
- A rectangle with a given perimeter which has the maximal area is a square
- A farmer planning to fence a rectangular garden to enclose the maximal area
- A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
- Finding the maximum area of the window of a special form
- OVERVIEW of lessons on finding the maximum/minimum of a quadratic function