Let s be the side of the square base and h be the height of the tank. Since the tank volume is restricted to 5324 ft cubed we have the following equation:

$V = s^2h = 5324$

$h = 5324 / s^2$

As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank

Base area with open top $s^2$

Side area 4sh

Total surface area $A = s^2 + 4sh$

We can substitute $h = 5324 / s^2$

$A = s^2 + 4s\frac{5324}{s^2}$

$A = s^2 + 21296/s$

To find the minimum of this function, we can take the first derivative, and set it to 0

$A' = 2s - 21296/s^2 = 0$

$2s = 21296/s^2$

$s^3 = 10648$

$s = \sqrt[3]{10648} = 22$

$h = 5324 / s^2 = 5324 / 22^2 = 11$

4 0

3 years ago

A store sells grape jelly in 16-ounces jars for $ 4.68. How much is the jelly per ounce?

guajiro [1.7K]

Answer:

.29 per ounce

Step-by-step explanation:

divided the cost by the amount

4.68/16

5 0

2 years ago

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