Answer:
r = 3.43 cm
h=18.94 cm
Step-by-step explanation:
The volume of the cylinder is:
[tex]V=\pi r^{2}h=700cm^{3}[/tex]
where r is the radius of the circular base and h is the height of the cylinder.
The cost of the side of the cylinder is:
[tex](0.02)2\pi rh[/tex]
The cost ot the bottom of the container is:
[tex](0.02)\pi r^{2}[/tex]
The cost ot the top of the container is:
[tex](0.09)\pi r^{2}[/tex]
Then, the total cost of the container is:
[tex]C=(0.02)2\pi rh+(0.02)\pi r^{2}+(0.09)\pi r^{2} \\C=\pi r(0.04h+0.02r+0.09r)\\C=\pi r(0.04h+0.11r)[/tex]
From the volume, you could solve for one variable and substituting in the cost equation:
[tex]h=\frac{700}{\pi r^{2}}\\C=\pi r(0.04(\frac{700}{\pi r^{2}})+0.11r)[/tex]
In order to minimize this fuction you need to calculate the derivative respect to r:
[tex]\frac{dC}{dr} =\frac{dC}{dr}[\frac{28}{r}+0.11\pi r^{2}]\\\frac{dC}{dr} =-\frac{28}{r^{2} }+0.22\pi r[/tex]
The critical points of the function are obtained when dC/dr=0:
[tex]-\frac{28}{r^{2}}+0.22\pi r=0\\0.22\pi r=\frac{28}{r^{2} }\\r^{3}= \frac{28}{0.22\pi } \\r=3.43 cm[/tex]
To evaluate if this critical point is a minimum, you should get the second derivative:
[tex]\frac{d^{2} C}{dr^{2} } =\frac{28}{r^{3} }+0.22\pi \\[/tex]
For the critical point r = 3.43 cm, the second derivative is positive, which means that the critical point is a minimum.
Then, the dimensions for the package tht will minimize product cost are:
r = 3.43 cm
[tex]h=\frac{700}{\pi r^{2} } =18.94cm[/tex]
