Find the equation of the line through point (3,−3) and parallel to y=3x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/
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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.