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The solution by "josgarithmetic" is .
Below find the correct solution.
The area of the garden, under the given condition, is
A = y*(80-2y) = -2y^2 + 80y.
Referring to the general form of a quadratic function
A = ay^2 + by +c,
the function have a maximum at y = , which is y = = = 20.
Thus the maximum is achieved at y = 20 m, and the maximal value of the quadratic function (of the area) is
A = -2*20^2 + 80*20 = -2*400 + 1600 = 800 square meters.
Answer. The dimensions of the garden are 20 m x 40 m. Its area is 800 square meters.
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The statement
At given perimeter, the area of a rectangle is maximal if and only if the rectangle is a square
is true if the PERIMETER (the entire perimeter consisting of four sides) is constrained.
In the given problem we have ANOTHER/DIFFERENT situation. (with which "josgarithmetic" is unfamiliar, due to his mathematical illiteracy).
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Plot A =
On finding the maximum/minimum of a quadratic function 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
- 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 farmer planning to fence a rectangular area along the river to enclose the maximal area (*)
- A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
- Using quadratic functions to solve problems on maximizing revenue/profit
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".
In the list of lessons, one is marked by the (*) sign.
It is your prototype/sample.
H a p p y l e a r n i n g ! !