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3 years ago

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An outdoor track consists of a rectangular region with a semi-circle on each end. If the perimeter of the track must be 200 mete

rs, find the dimensions that will make the area of the rectangular region as large as possible.

1 answer:

mars1129 [50]3 years ago

6 0

Answer:

Length = 50m

Width = 31.84m

Step-by-step explanation:

Perimeter of a track = 200m

The perimeter of a track with two semi-circular end and two rectangular regions = P

Perimeter of a semi-circle = ½(2πr)

Length of a rectangle = x

Perimeter of the rectangle = x+x

P = ½(2πr) + ½(2πr)+x + x

P = 2πr + 2x

200 =2πr + 2x

2πr = 200 – 2x

2πr = 2(100 – x)

r = 2(100 – x) / 2π

r = (100-x)/π

total area of the rectangular region =A

A= x(2r)

= (x)2[(100-x)/π]

= 2x[(100-x)/π]

A = (200x – 2x^2)/π

Differentiate A with respect to x

dA/dx = (200 -4x)/π

at critical point, first derivative vanishes(dA/dx = 0)

(200 -4x)/π = 0

-4x =-200

x =-200/-4

x = 50

this means length (x) = 50m

put x=50 into r = (100-x)/π

r= (100-50)/π

r =50/π

width = 2r

width = 2(50/π)

= 2(19.62)

= 31.84m

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$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}$ .

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$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$ .

(Round to at least one more decimal places than the values in the choices.)

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$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}$ .

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}$ .

$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$ .

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