Hello someone tell me what I should do. Don't give answer just tell me.
1 answer:
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Let h = height of the box,
x = side length of the base.
Volume of the box is
.
So
Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.
![4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86](https://tex.z-dn.net/?f=%204x%20%3D%20%5Cfrac%7B460%7D%7B%20x%5E%7B2%7D%20%7D%20%20%5C%5C%20%204x%5E%7B3%7D%20%3D%20460%20%20%5C%5C%20x%5E%7B3%7D%20%3D%20115%20%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B115%7D%20%3D%204.86%20)
Then
So the box is 4.86 in. wide and 4.87 in. high.
x = side length of the base.
Volume of the box is
So
Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.
Then
So the box is 4.86 in. wide and 4.87 in. high.
In an hour the minute hand goes all the way around, or a 360 loop. so it's its 
times the radius squared.