Respuesta :
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:
⇒ [tex]P(r)=(unit\ cost)\ (SA) - (unit\ cost)\ (Volume)[/tex]
⇒ [tex]P(r)=(30)\ (4 \pi r^2) - (500)\ (\frac{4\pi r^3}{3} )[/tex]
⇒ [tex]P(r)=(120)\ (2\pi r^2) - (\frac{500\times 4}{3} )\ \pi r^3[/tex]
⇒ [tex]P(r)=(120)\ (\pi r^2) - (\frac{2000}{3} )\ \pi r^3[/tex]
Differentiate "[tex]P[/tex]" and find the "[tex]r[/tex]" value then double differentiate "[tex]P[/tex]", plug the "[tex]r[/tex]" values from [tex]P'[/tex] to find the minimum and maximum values.
⇒ [tex]P(r)'=(120)\ 2\pi r - (\frac{2000}{3} )\ 3\pi r^2[/tex]
⇒ [tex]P(r)'=(240)\ \pi r - (2000)\ \pi r^2[/tex]
Finding r values :
⇒ [tex](240)\ \pi r - (2000)\ \pi r^2 =0[/tex] Â
Dividing both sides with 240Ï€ .
⇒ [tex]r-\frac{25}{3} r^2 =0[/tex]   ⇒ [tex]r(1-\frac{25}{3} r) =0[/tex]
⇒ [tex]r=0[/tex] and [tex]r=\frac{3}{25}[/tex] Â
To find maxima value the double differentiation is :
⇒ [tex]P(r)'=(240)\ \pi r - (2000)\ \pi r^2[/tex]   ...first derivative
Double differentiating :
⇒ [tex]P(r)''=(240\pi) - (2000\pi)\ 2(r)[/tex]   ...second derivative
⇒ [tex]P(r)''=(240\pi) - (4000\pi)\ (r)[/tex]
Test the value r = 3/25 dividing both sides with 240Ï€
⇒  [tex]1 - \frac{50\pi r}{3}[/tex]
⇒ [tex]1 - \frac{50\times \pi\times 3 }{3\times 25}[/tex]
⇒ [tex]-5.28 < 0[/tex]
It passed the double differentiation test.
Extra work :
Thus:
⇒ [tex]P(r)=(120)\ (\pi r^2) - (\frac{2000}{3} )\ \pi r^3[/tex]
⇒ [tex]P(r)=(120)\times (\pi (\frac{3}{25} )^2) - (\frac{2000}{3} )\times \pi (\frac{3}{25} )^3[/tex]
⇒ [tex]P(r) =1.8095[/tex]
Finally r =3/25 ft that will maximize the profit of the manufacturing company.
