Among all rectangles having a perimeter of 25 m, find the dimensions of the one with the largest area. (help :()?
回答 (5)
2015-10-06 12:44 pm
Length = x, width = 12.5 - x {perimeter = 25} &
area,A, = 12.5x = x^2, (dA/dx) = A' = 12.5 - 2x &
A'' = - 2--> a has rel max @x = 6.25. Then max
area is in a rectangle having length=width=6.25.
So rectangle is a square of side length 6.25 m.
area,A, = 12.5x = x^2, (dA/dx) = A' = 12.5 - 2x &
A'' = - 2--> a has rel max @x = 6.25. Then max
area is in a rectangle having length=width=6.25.
So rectangle is a square of side length 6.25 m.
2015-10-06 7:36 am
If w represents the width of the rectangle, 12.5 - w represents the length. Then area is a quadratic function of w
.. area = w(12.5-w)
This has a vertex halfway between the zeros 0 and 12.5, at w=6.25, the dimension that makes the rectangle a square.
The rectangle of perimeter 25 m having the largest area is a square of side length 6.25 m.
.. area = w(12.5-w)
This has a vertex halfway between the zeros 0 and 12.5, at w=6.25, the dimension that makes the rectangle a square.
The rectangle of perimeter 25 m having the largest area is a square of side length 6.25 m.
2015-10-06 6:30 am
The area of a rectangle is length x width, so first find all the combinations of 4 numbers that add up to 25. The combination that gives you the two different larger number will be your answer.
Therefore, the rectangle with the largest area would be 7.5x5.
I am assuming they are purposely saying a rectangle. Unless a square is acceptable, in which case just divide 25 by 4 :)
Therefore, the rectangle with the largest area would be 7.5x5.
I am assuming they are purposely saying a rectangle. Unless a square is acceptable, in which case just divide 25 by 4 :)
2015-10-06 6:28 am
The answer is a square of side length 25/4, but if you need to do it with calculus:
P = 2L + 2W
W = (P - 2L) / 2
A = L x W = L(P - 2L) / 2 = (P/2)L - L²
dA/dL = (P/2) - 2L = 0
2L = P/2
L = P/4
W = (P - 2(P/4)) / 2 = (P - (P/2)) / 2 = P/4
So L = W = P/4 = 25/4.
P = 2L + 2W
W = (P - 2L) / 2
A = L x W = L(P - 2L) / 2 = (P/2)L - L²
dA/dL = (P/2) - 2L = 0
2L = P/2
L = P/4
W = (P - 2(P/4)) / 2 = (P - (P/2)) / 2 = P/4
So L = W = P/4 = 25/4.
2015-10-06 6:28 am
it will be a square...to show : A = L W and L + W = 25/2.....L= 12.5 - W into the area , maximize the area function
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