The area of a rectangle whose perimeter is a fixed 80 feet is given by A=40w - w2 , where w is the width of the rectangle. Determine the width of the rectangle that gives the maximum area. What type of special rectangle is necessary to produce this maximum area? Justify.
Width=20 feet
Since Length=Width=20 feet, the rectangle is a Square.
Step-by-step explanation:
Area, [tex]A=40w - w^2[/tex]
To determine the width of the rectangle that gives the maximum area, we take the derivative of A and solve for its critical point.
[tex]A'=40 - 2w\\$When A'=0\\40-2w=0\\40=2w\\w=20 feet[/tex]
The width of the rectangle that gives the maximum area =20 feet.
Perimeter of a rectangle=2(l+w)
Perimeter of the rectangle=80 feet
2(l+w)=80
2l+2(20)=80
2l=80-40
2l=40
l=20 feet
Since the length and width are equal, the special type of rectangle that produces this maximum area is a Square.
something over 10 so ? /10
step-by-step explanation:
25 questions , because they start with 100.
350-100=250 .
. 250/10=25
step-by-step explanation:
first take off 100, then what x 10 = 250? that is your answer. (25)