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Tomtit [17]
2 years ago
12

One-stray Problem Solving

Mathematics
1 answer:
anygoal [31]2 years ago
3 0
A) 5000 m²
b) A(x) = x(200 -2x)
c) 0 < x < 100
Step-by-step explanation:
b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
A(x) = x(200 -2x)
__
a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
__
c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.
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Answer:

For maximum area, all of the wire should be used to construct the square.

The minimum total area is obtained when length of the wire is 10m

Step-by-step explanation:

For maximum,  we use the whole length

For minimum,

supposed the x length was used for the square,

the length of the side of the square = x/4m

Area = \frac{x^{2} }{16}

For the equilateral triangle, the length of the side =  \frac{23 - x}{3}

Area = \frac{\sqrt{3} }{4}  a^{2}  = \frac{\sqrt{3} }{4} (\frac{23 - x}{3} )^{2}

Total Area = \frac{x^{2} }{16}  + \frac{\sqrt{3} }{36} (23-x)^{2}

\frac{dA}{dx}  = \frac{x}{8}  -  \frac{\sqrt{3} }{18} (23 - x)\\

\frac{d^{2}A }{dx^{2} }  = \frac{1}{8}  + \frac{\sqrt{3} }{18}  > 0, therefore it is minimum

\frac{dA}{dx}  = 0 \\\\

\frac{x}{8}  -  \frac{\sqrt{3} }{18} (23 - x) = 0\\

x = 10.00m

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