Answer:
The value of r to have maximum profit is 3/25 ft
Step-by-step explanation:
To find:
The size of the sphere so that the profit can be maximized.
Manufacturing cost of the solid sphere = $500/ ft^3
Selling price of sphere (on surface area) = $30 / ft^2
We see that the manufacturing cost dealt with he volume of the sphere where as the selling price dealt with the surface area.
So,
To maximize the profit (P) .
We can say that:
⇒ ![P(r)=(unit\ cost)\ (SA) - (unit\ cost)\ (Volume)](https://tex.z-dn.net/?f=P%28r%29%3D%28unit%5C%20cost%29%5C%20%28SA%29%20-%20%28unit%5C%20cost%29%5C%20%28Volume%29)
⇒ ![P(r)=(30)\ (4 \pi r^2) - (500)\ (\frac{4\pi r^3}{3} )](https://tex.z-dn.net/?f=P%28r%29%3D%2830%29%5C%20%284%20%5Cpi%20r%5E2%29%20-%20%28500%29%5C%20%28%5Cfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D%20%29)
⇒ ![P(r)=(120)\ (2\pi r^2) - (\frac{500\times 4}{3} )\ \pi r^3](https://tex.z-dn.net/?f=P%28r%29%3D%28120%29%5C%20%282%5Cpi%20r%5E2%29%20-%20%28%5Cfrac%7B500%5Ctimes%204%7D%7B3%7D%20%29%5C%20%5Cpi%20r%5E3)
⇒ ![P(r)=(120)\ (\pi r^2) - (\frac{2000}{3} )\ \pi r^3](https://tex.z-dn.net/?f=P%28r%29%3D%28120%29%5C%20%28%5Cpi%20r%5E2%29%20-%20%28%5Cfrac%7B2000%7D%7B3%7D%20%29%5C%20%5Cpi%20r%5E3)
Differentiate "
" and find the "
" value then double differentiate "
", plug the "
" values from
to find the minimum and maximum values.
⇒ ![P(r)'=(120)\ 2\pi r - (\frac{2000}{3} )\ 3\pi r^2](https://tex.z-dn.net/?f=P%28r%29%27%3D%28120%29%5C%202%5Cpi%20r%20-%20%28%5Cfrac%7B2000%7D%7B3%7D%20%29%5C%203%5Cpi%20r%5E2)
⇒ ![P(r)'=(240)\ \pi r - (2000)\ \pi r^2](https://tex.z-dn.net/?f=P%28r%29%27%3D%28240%29%5C%20%5Cpi%20r%20-%20%282000%29%5C%20%5Cpi%20r%5E2)
Finding r values :
⇒
Dividing both sides with 240π .
⇒
⇒ ![r(1-\frac{25}{3} r) =0](https://tex.z-dn.net/?f=r%281-%5Cfrac%7B25%7D%7B3%7D%20r%29%20%3D0)
⇒
and
To find maxima value the double differentiation is :
⇒
...first derivative
Double differentiating :
⇒
...second derivative
⇒ ![P(r)''=(240\pi) - (4000\pi)\ (r)](https://tex.z-dn.net/?f=P%28r%29%27%27%3D%28240%5Cpi%29%20-%20%284000%5Cpi%29%5C%20%28r%29)
Test the value r = 3/25 dividing both sides with 240π
⇒
⇒ ![1 - \frac{50\times \pi\times 3 }{3\times 25}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B50%5Ctimes%20%5Cpi%5Ctimes%203%20%7D%7B3%5Ctimes%2025%7D)
⇒ ![-5.28 < 0](https://tex.z-dn.net/?f=-5.28%20%3C%200)
It passed the double differentiation test.
Extra work :
Thus:
⇒ ![P(r)=(120)\ (\pi r^2) - (\frac{2000}{3} )\ \pi r^3](https://tex.z-dn.net/?f=P%28r%29%3D%28120%29%5C%20%28%5Cpi%20r%5E2%29%20-%20%28%5Cfrac%7B2000%7D%7B3%7D%20%29%5C%20%5Cpi%20r%5E3)
⇒ ![P(r)=(120)\times (\pi (\frac{3}{25} )^2) - (\frac{2000}{3} )\times \pi (\frac{3}{25} )^3](https://tex.z-dn.net/?f=P%28r%29%3D%28120%29%5Ctimes%20%28%5Cpi%20%28%5Cfrac%7B3%7D%7B25%7D%20%29%5E2%29%20-%20%28%5Cfrac%7B2000%7D%7B3%7D%20%29%5Ctimes%20%5Cpi%20%28%5Cfrac%7B3%7D%7B25%7D%20%29%5E3)
⇒
Finally r =3/25 ft that will maximize the profit of the manufacturing company.
Step-by-step explanation:
Adhala mudiyadhu kelambu vandhutha eva
Answer:
34.8 N at 2.5 degrees from the positive x-axis
Step-by-step explanation:
From the given information:
The force F makes an angle A with the positive x axis can be expressed in terms horizontal and vertical components.
![F_x = F cos A\\\\ F_y = Fsin A](https://tex.z-dn.net/?f=F_x%20%3D%20F%20cos%20A%5C%5C%5C%5C%20F_y%20%3D%20Fsin%20A)
Given that
![F_1 = 50 \ N](https://tex.z-dn.net/?f=F_1%20%3D%2050%20%5C%20N)
![\theta_1 = 30 ^0 \ \ \ (x-axis)](https://tex.z-dn.net/?f=%5Ctheta_1%20%3D%2030%20%5E0%20%5C%20%20%5C%20%20%5C%20%28x-axis%29)
![F_{1x} = F_1 \times Cos A_1](https://tex.z-dn.net/?f=F_%7B1x%7D%20%3D%20F_1%20%5Ctimes%20Cos%20A_1)
= 50 × cos 30
= 43.3 N
![F_{1y} = F_1 \times Sin \ A_1](https://tex.z-dn.net/?f=F_%7B1y%7D%20%3D%20F_1%20%5Ctimes%20Sin%20%5C%20A_1)
= 50 × sin 30
= 25 N
Similarly;
![F_2 = 25 \ N](https://tex.z-dn.net/?f=F_2%20%3D%2025%20%5C%20N)
![\theta_2 = 250 ^0 \ \ \ ( x-axis)](https://tex.z-dn.net/?f=%5Ctheta_2%20%3D%20250%20%5E0%20%5C%20%5C%20%5C%20%28%20x-axis%29)
![F_{2x} = F_2 \times cos \ A_2](https://tex.z-dn.net/?f=F_%7B2x%7D%20%3D%20F_2%20%5Ctimes%20cos%20%20%5C%20A_2)
= 25 × cos 250
= - 8.55 N
![F_{2y} = F_2 \times A_2](https://tex.z-dn.net/?f=F_%7B2y%7D%20%3D%20F_2%20%5Ctimes%20A_2)
= 25 × sin 250
= -23.5 N
The total net force;
![F_{net} = F_{(net)}_x + F_{(net)}_y](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F_%7B%28net%29%7D_x%20%2B%20F_%7B%28net%29%7D_y)
![F_{net} = (F_{1x} + F_{2x} ) i + (F_{1y} + F_{2y} )j](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%28F_%7B1x%7D%20%2B%20F_%7B2x%7D%20%29%20i%20%2B%20%28F_%7B1y%7D%20%2B%20F_%7B2y%7D%20%29j)
![F_{net} = (43.3 - 8.55) i + (25-23.5 ) j](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%2843.3%20-%208.55%29%20i%20%2B%20%2825-23.5%20%29%20j)
![F_{net} =34.75 i + 1.5 j](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D34.75%20i%20%2B%201.5%20j)
![|F_{net} | = \sqrt{F_{net}_x^2 + F_{net}_y^2}](https://tex.z-dn.net/?f=%7CF_%7Bnet%7D%20%7C%20%3D%20%5Csqrt%7BF_%7Bnet%7D_x%5E2%20%2B%20F_%7Bnet%7D_y%5E2%7D)
![|F_{net} | = \sqrt{34.75^2 + 1.5^2}](https://tex.z-dn.net/?f=%7CF_%7Bnet%7D%20%7C%20%3D%20%5Csqrt%7B34.75%5E2%20%2B%201.5%5E2%7D)
![|F_{net} | = 34.8 \ N](https://tex.z-dn.net/?f=%7CF_%7Bnet%7D%20%7C%20%3D%2034.8%20%5C%20N)
Finally, the direction of the angle for the net force is:
![tan \theta = \dfrac{F_{net_y}}{F_{net_x}}](https://tex.z-dn.net/?f=tan%20%5Ctheta%20%3D%20%5Cdfrac%7BF_%7Bnet_y%7D%7D%7BF_%7Bnet_x%7D%7D)
![\theta = tan^{-1} \Big (\dfrac{F_{net_y}}{F_{net_x}} \Big)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5CBig%20%28%5Cdfrac%7BF_%7Bnet_y%7D%7D%7BF_%7Bnet_x%7D%7D%20%5CBig%29)
![\theta = tan^{-1} \Big (\dfrac{1.5}{34.75}} \Big)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5CBig%20%28%5Cdfrac%7B1.5%7D%7B34.75%7D%7D%20%5CBig%29)
![\theta = tan^{-1}( 0.043165)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%28%200.043165%29)
![\theta \simeq 2.5^0\ with \ positive \ x-axis](https://tex.z-dn.net/?f=%5Ctheta%20%5Csimeq%202.5%5E0%5C%20%20with%20%5C%20positive%20%5C%20x-axis)
By direct division, we will find that the length of each piece of rope is 10.5m
<h3>
How to perform the division?</h3>
So e start with a 42 meter long rope and we cut it into 4 equal parts, then we know that each one of these pieces will have a length equal to:
L = (42m/4)
To perform that division, you can see that 42<u> is not a multiple of 4</u>, so we can write:
L = (40m + 2m)/4 = (40m/4) + (2m/4)
The first quotient is just equal to 10m, while the second one is:
(2m/4) = (1m/2) = 0.5m
Then we have:
L = (40m/4) + (2m/4) = 10m + 0.5m = 10.5m
Each piece of rope has a length of 10.5m
If you want to learn more about divisions, you can read:
brainly.com/question/7068223
Answer:
Step-by-step explanation:
11 - 19 - 4 - (-8) =
11 - 19 - 4 + 8 =
- 23 + 19 =
- 4 <===