.
x-coordinate of the point P is "u";
y--coordinate of the point P is "v".
These coordinates satisfy equation
u + v = 4.
The area of the triangle is
S = = = = .
Thus the area "S" is the quadratic function of "u".
It achieves the maximum value at u = " " = = = 2.
The value of the maximum is S = = = = 2.
ANSWER. The area of the triangle has the maximum at the point P = (2,2), and the value of the maximum is 2.
Solved.
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If you want to expand your knowledge in this class of problem and if you want to feel a solid ground under your legs,
look into 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
- OVERVIEW of lessons on finding the maximum/minimum of a quadratic function
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.