.
Let x be the length in yards of the front fence (at 15 dollars per yard).
Then the other dimension will be yards, or yards = (45 - 3x) yards.
The area is the product of these dimensions
Area A = square yards.
And the problem asks to find the maximal area as the maximum of this quadratic form.
This quadratic form has the zeroes 0 and = 15, so its maximum is achieved exactly at the midpoint between the zeroes, i.e at x= 7.5.
Answer. Under the given condition, the maximum area is the rectangle with one dimension 7.5 yards (front yard at $15/yard)
and other dimension 45-3*7.5 = 22.5 yards.
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On finding minimum/maximum of quadratic functions and solving similar problems 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
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".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.