Respuesta :
Answer:
Radius =6.518 feet
Height = 26.074 feet
Step-by-step explanation:
The Volume of the Solid formed = Volume of the two Hemisphere + Volume of the Cylinder
Volume of a Hemisphere [tex]=\frac{2}{3}\pi r^3[/tex]
Volume of a Cylinder [tex]=\pi r^2 h[/tex]
Therefore:
The Volume of the Solid formed
[tex]=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}[/tex]
Area of the Hemisphere =[tex]2\pi r^2[/tex]
Curved Surface Area of the Cylinder =[tex]2\pi rh[/tex]
Total Surface Area=
[tex]2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh[/tex]
Cost of the Hemispherical Ends = 2 X Cost of the surface area of the sides.
Therefore total Cost, C
[tex]=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh[/tex]
Recall: [tex]h=\frac{4640}{\pi r^2}-\frac{4r}{3}[/tex]
Therefore:
[tex]C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}[/tex]
The minimum cost occurs at the point where the derivative equals zero.
[tex]C^{'}=\frac{-27840+32\pi r^3}{3r^2}[/tex]
[tex]When \:C^{'}=0[/tex]
[tex]-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518[/tex]
Recall:
[tex]h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet[/tex]
Therefore, the dimensions that will minimize the cost are:
Radius =6.518 feet
Height = 26.074 feet
The dimensions that minimize the cost of building the tank are [tex]r\approx 6.5ft \text{ and } h \approx 26.0ft[/tex] .
Let
[tex]h=\text{the height of the cylindrical sides}\\r=\text{the radius of the hemispheres and the cylindrical sides}[/tex]
and
[tex]T=\text{the total cost}\\u_c=\text{the unit cost of building the cylindrical sides}\\u_h=\text{the unit cost of building the hemispherical surfaces}[/tex]
and
[tex]S_c=\text{the surface area of the cylinder}\\=2\pi rh\\\\S_h=\text{the total surface area of both hemispheres}\\=4\pi r^2[/tex]
Therefore,
[tex]T=u_cSc+u_hSh[/tex]
From the question
[tex]u_h=2u_c[/tex]
The total cost becomes
[tex]T=u_cSc+2u_cSh\\=2u_c \pi rh+8u_c \pi r^2[/tex]
We need to eliminate [tex]h[/tex]. The volume from the question gives a way out
[tex]4640=\frac{4}{3}\pi r^3 +\pi r^2h\\\\h=\frac{4640}{\pi r^2}-\frac{4r}{3}[/tex]
substitute into the formula for total cost gives, after simplifying
[tex]T=\frac{16u_c\pi r^2}{3}+\frac{9280u_c}{r^2}[/tex]
differentiating with respect to [tex]r[/tex], we get
[tex]\frac{dT}{dr}=\frac{32}{3}u_c\pi r-\frac{9280u_c}{r^2}[/tex]
at extrema
[tex]\frac{dT}{dr}=0\\\\\implies \frac{32}{3}u_c\pi r=\frac{9280u_c}{r^2}\\\\r=\sqrt[3]{\frac{870}{\pi}}\approx 6.5ft[/tex]
To confirm that [tex]r[/tex] is a minimum value, carry out the second derivative test
[tex]\frac{d^2T}{dr^2}=\frac{32u_c\pi}{3}+\frac{18560}{r^3}[/tex]
substituting [tex]r=\sqrt[3]{\frac{870}{\pi}}[/tex], we get that [tex]\frac{d^2T}{dr^2}[/tex] > [tex]0[/tex], confirming that minimum value
To find [tex]h[/tex], recall that
[tex]h=\frac{4640}{\pi r^2}-\frac{4r}{3}[/tex]
substituting [tex]r[/tex], we get [tex]h\approx 26.0ft[/tex] as the corresponding minimum height
Therefore, [tex]r\approx 6.5ft \text{ and } h \approx 26.0ft[/tex] minimize the total cost of building the tank.
Learn more about minimizing dimensions to reduce costs here: https://brainly.com/question/19053049
