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almond37 [142]
3 years ago
15

(1) Logan has 6 pounds of paste. Each time he makes dinner he uses 34 pound of

Mathematics
1 answer:
kaheart [24]3 years ago
3 0

SHDHSB SSHD EUDXD EJDJXJEX SJDJWIDIJDWN SSHWIJJSJ

Step-by-step explanation:

XJDUXIDB DUDUU

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Find the measurements (the length L and the width W) of an inscribed rectangle under the line with the 1st quadrant of the x &am
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

To maximize, we have to differentiate the equation:

\frac{dA}{dx} = \frac{d}{dx}(-\frac{3}{4}.x^{2}  + 3x)

\frac{dA}{dx} = -3x + 3

The critical point is:

\frac{dA}{dx} = 0

-3x + 3 = 0

x = 1

Substituing:

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

The maximum area is:

A = 1 . 9/4

A = 9/4

A = 2.25

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3 years ago
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