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Fiesta28 [93]
3 years ago
13

He length of a rectangle is three times its width. The perimeter of the rectangle is at most 112 cm.

Mathematics
2 answers:
g100num [7]3 years ago
8 0
The equation relating length to width
L = 3W
The inequality stating the boundaries of the perimeter
LW <= 112
When you plug in what L equals in the first equation into the second equation, you get
3W * W <= 112
evaluate
3W^2 <= 112
3W <= 4 \sqrt{7}
W <= \frac{4 \sqrt{7} }{3} cm
Semenov [28]3 years ago
8 0

Answer:

the answer is 2w+2*(3w)≤112

Step-by-step explanation:

If the rectangle is three times the width and the perimeter is 112 cm then we do 2w+2*(3w)≤112 since it is bigger or the same as 112

btw the other guy is so wrong

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Leni [432]

The question is incomplete. Here is the complete question.

Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -\frac{3}{4}x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.

Answer: L = 1; W = 9/4; A = 2.25;

Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:

A = x.y

A = x(-\frac{3}{4}.x + 3)

A = -\frac{3}{4}.x^{2}  + 3x

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\frac{dA}{dx} = 0

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y = -\frac{3}{4}x + 3

y = -\frac{3}{4}.1 + 3

y = 9/4

So, the measurements are x = L = 1 and y = W = 9/4

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A = 1 . 9/4

A = 9/4

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