SOLUTION: The perimeter of a rectangle remains fixed at 16cm. By first writing, an equation involving the sides of the rectangle, find the dimensions that give the maximum area.
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Question 1120518: The perimeter of a rectangle remains fixed at 16cm. By first writing, an equation involving the sides of the rectangle, find the dimensions that give the maximum area.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Sides are x and 8-x, to make a half perimeter of 8. 2x+16-2x=16
Maximize the product of those 2
x(8-x)=-x^2+8x
Using calculus, first derivative is -2x+8=0 or 2x=8 and x=4, a square 4 x 4 with area 16
Or, parabola with vertex x value at -b/2a which occurs at -8/-2 or 4, and the same result.
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
At given perimeter P of a rectangle, the greatest area is achieved when the rectangle is a square
(and then its side is, obviously, P/4 units of length).
It is a general fact.
For algebraic proof and discussion (and examples of solved problems) see the lesson
- A rectangle with a given perimeter which has the maximal area is a square
in this site.
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
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.
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