What dimensions should the garden be to maximize the area while keeping the same perimeter? - dimensions of a pinewood derby block of wood
The garden is 160 square meters and the circumference is 52 feet
How do I do that? I know the dimensions of the garden are 10 × 16 meters, but what does "maximize the area while retaining the same extent, and how to solve them?
Tuesday, January 26, 2010
Dimensions Of A Pinewood Derby Block Of Wood What Dimensions Should The Garden Be To Maximize The Area While Keeping The Same Perimeter?
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3 comments:
Under the assumption that the garden is a square, the question is why the two dimensions, length and width are multiplied together to give the largest area.
The best way is simply given by the size of two. This would give 26 in this case. Then on the multiplication of two numbers that are found to be 26 until you have, which gives children the most product categories. In this case I think the answer would have dimensions of 13 x 13, ie, the garden is a square. What does that mean?
I am new to this answers all questions, so I hope this helps. haha
Perimeter is determined by the function
p = 2W 2 L
52 = 2W 2 L
is given by the function
A = lw
want maximum area given perimeter
52 = 2W 2 L
(52-2V) / 2 = L
maxarea w = (52-2V) / 2
maxarea = (52W-2W ^ 2) / 2
maxarea = 26W-w ^ 2
When you do the algebra, you put it to use in your graphing calculator, and most often. other wise take the deduction. maxarea 0 is, by the derviative
0 = 26-2W
-26 =- 2w
w = 13
52 = 2 (13) 2 L
26 = 2L
13 = L
Therefore, the length and width is to be 13feet, to an area of 169 m ^ 2
If it is a square or a rectangle? If so, it maximizes the space, if there is a place so that the parties are both 13 meters and the region of 169 m ^ 2 would be.
If you fit the edges, then the surface will be maximized if it is a circle:
perimeter = 52 = 2 (pi) (R), so that r 8.28. Then the surface (pi) (R) ^ 2 = (3.14) (8.28) ^ 2 = 215 m ^ 2 is
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