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Mathematics

1 answer:

Sholpan [36]3 years ago

4 0

Let
x-------> the width of the rectangular area
y------> the length of the rectangular area

we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2

substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft

see the attached figure
the solution is<span> the shaded area</span>

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Find the area of the circle x^2+y^2=16 by the method of intregration

zhuklara [117]

Answer:

Hello,

$16\pi$

Step-by-step explanation:

$I=\dfrac{Area}{4} =\int\limits^4_0 {\sqrt{16-x^2} } \, dx \\\\Let\ say\ x=4*sin(t),\ dx=4*cos(t) dt\\\\\displaystyle I=\int\limits^\frac{\pi }{2} _0 {4*\sqrt{1-sin^2(t)} }*4*cos(t) \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {cos^2(t)} \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {\frac{1-cos(2t)}{2}} \, dt \\\\=8*[t]^\frac{\pi }{2} _0-[\frac{sin(2t)}{2} ]^\frac{\pi }{2} _0\\\\=4\pi -0\\\\=4\pi\\\\\boxed{Area=4*I=16\pi}\\$