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RafaP = 2(L + W)
50 = 2(L + W)
25 = L + W
Let L = x, so W = 25 - x
A = LW
A = x(25 - x)
A = -x2 + 25x
Since the parabola that represents the above equation opens downward, we have a maximum point (the y-coordinate value of the vertex of the parabola, gives us the maximum value of the area).
vertex x-coordinate value = - b/2a = - 25/-2 = 25/2
vertex y-coordinate value = -(25/2)2 + 25(25/2) = - 252/4 + 252/2 = - 252/4 + 2(252)/4 = 252/4 = 156.25
Thu, the maximum area will be 156.25 unit2.
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What is the perimeter of 50 yards and 100 yards?
The perimeter of a rectangle of dimensions 50 yards x 100 yards is 300 yards.
Find the perimeter of a rectangle with the length of 15and the width of 10?
The formula to calculate perimeter of a rectangle is P=2L+2W P= 2(15) + 2(10) P= 30 +20 P = 50
The length of a rectangle is twice its width If the area of the rectangle is less than 50 square meters what is the greatest width of the rectangle?
From the statement of the problem, if w is the width, the area is 2w2 , the product of the width and the length, which is stated to be twice the width. Since 2w2 must be less than 50, w2 < 25, and the width must be less than 5 meters.
The length of a rectangle is twice its width If the area of the rectangle is 50yds2 What is the perimeter?
Answer: 30 yards. Explanation: let's say it is L yards long and W yards wide. Then perimeter is 2L+2W The area is LxW=50 So Since LW=50, we substitute L=2W and write 2W(W)=2W^2=50. so W^2=50/2=25 and W=5 Now 2W=10 which is L So is it 10 yards long and 5 yards wide Let's check that. Area is 10x5=50 as desired and Length is 10 which is twice 5 as desired Now perimeter is 2L+2W=2(10)+2(5)=20+10=30 yards.
The perimeter of a rectangle is 130m. The length of the rectangle is 25m more than the width. What are the lengths and widths of the rectangle?
x=width x+25=length 4x+50=130 4x=80 x=20 width is 20 length is 45
