Answer:
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Step-by-step explanation:
Answer:
- 2(L +W) ≤ 600
- W ≤ 200
- L ≥ 2W
Step-by-step explanation:
We assume the problem wording means the length is to be at least 2 times <em>as long as</em> the width. (<em>Longer than</em> usually refers to a difference, not a scale factor.)
If we let "W" and "L" represent the width and length, respectively, then we can translate the problem statement to ...
2(L + W) ≤ 600 . . . . . . the perimeter is twice the sum of length and width
W ≤ 200 . . . . . . . . . . . . the width is at most 200 inches
L ≥ 2W . . . . . . . . . . . . . the length is at least twice the width
If A=w(50-w)
A=50w-w^2
dA/dw=50-2w
d2A/dw2=-2
Since the acceleration is a constant negative, that means that when velocity, dA/dw=0, it is at an absolute maximum for A(w)...
dA/d2=0 only when 50=2w, w=25
So as the case with any rectangle, the perfect square will enclose the greatest area possible with respect to a given amount of material to enclose that area...
So the greatest area occurs when W=L=25 in this case:
A(25)=50w-w^2
Area maximum is thus:
Amax=50(25)-(25)^2=625 u^2
Answer:
Length=6
Width=3
Step-by-step explanation:
Area:L*W
Area= 6*3=18
Perimeter: 2 (L+W)
Perimeter= 2 (6+3)= 12+6=18
The answer is 140.07< 140.7 I hope I help
