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3 years ago

Please please help me and fast

Mathematics

1 answer:

Eduardwww [97]3 years ago

4 0

Its 12. Part a was 2 part b was 3
6×2=12
4×3=12

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Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

$h=\frac{400}{\pi r^2}$

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: $h=\frac{400}{\pi r^2}$

Therefore:

$C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})$

$C(r)=0.12\pi r^2+\frac{24}{r}$

$C(r)=\frac{0.12\pi r^3+24}{r}$

The minimum cost occurs when the derivative of the Cost =0.

$C^{'}(r)=\frac{6\pi r^3-600}{25r^2}$

$6\pi r^3-600=0$

$6\pi r^3=600$

$\pi r^3=100$

$r^3=\frac{100}{\pi}$

$r^3=31.83$

r=3.17 cm

Recall that:

$h=\frac{400}{\pi r^2}$

$h=\frac{400}{\pi *3.17^2}$

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

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