As part of a collage for her art class, Sheila wants to enclose a rectangle with 100 inches of yarn. Write an expression for the area A of the rectangle in terms of w (width).
B) write an expression for its length
C) what is the width of the rectangle she must make if Sheila want a rectangle that encloses the maximum area?
D) what is the area of the largest rectangle that Sheila can enclose with 100 inches of yam?
Answers
Answer
It is given that, Sheila wants to enclose a rectangle with 100 inches of yarn.
So the perimeter of rectangle is 100/2 = 50
Therefore Length + width = 50
Let 'w' be the width of rectangle.
Width = w
Length = 50 - w
A) Write an expression for the area A of the rectangle in terms of w (width).
Area, A = length x width
A = w(50 - w) = 50w -
B) write an expression for its length
Length = 50 - w
C) The width of the rectangle that encloses the maximum area
Sum of length and width is 50
The product of length and width become maximum when
length 50/2 +1 = 26 inches and
width 50/2 -1 = 24 inches
D) The area of the largest rectangle that Sheila can enclose with 100 inches of yam
Area = length x width = 26 x 24 = 624
Area = 624 square inches
Answer:
It is given that, Sheila wants to enclose a rectangle with 100 inches of yarn.
So the perimeter of rectangle is 100/2 = 50
Therefore Length + width=50
Let 'w' be the width of rectangle.
Width = w
Length = 50-w
A) Write an expression for the area A of the rectangle in terms of w (width).
Area, A = length x width
A = w(50-w) = 50w - w²
B) write an expression for its length
Length= 50-w
C) The width of the rectangle that encloses the maximum area
Sum of length and width is 50
The product of length and width become maximum when
length 50/2 +1 26 inches and
width 50/2 -1 = 24 inches
D) The area of the largest rectangle that Sheila can enclose with 100 inches of yam
Area = length x width = 26 x 24 = 624
Area = 624 square inches
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