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A cushion filled with stuffing. In geometry, the paper bag problem or teabag problem is to calculate the maximum possible inflated volume of an initially flat sealed rectangular bag which has the same shape as a cushion or pillow, made out of two pieces of material which can bend but not stretch.
Area#Area formulas – Size of a two-dimensional surface; Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities
Such a formula would be needed for building pyramids. In the next problem (Problem 57), the height of a pyramid is calculated from the base length and the seked (Egyptian for slope), while problem 58 gives the length of the base and the height and uses these measurements to compute the seked.
The formula for the volume of a pyramidal square frustum was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written in the 13th dynasty (c. 1850 BC): = (+ +), where a and b are the base and top side lengths, and h is the height.
The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m-1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus
It is also possible to approximate the minimum bounding box volume, to within any constant factor greater than one, in linear time. The algorithm for doing this involves finding an approximation to the diameter of the point set, and using a box oriented towards this diameter as an initial approximation to the minimum volume bounding box.
Problems of this type included finding the dimensions of a rectangle given its area and the amount by which the length exceeds the width. Tables of values of n 3 + n 2 were used to solve certain cubic equations. For example, consider the equation: + =. Multiplying the equation by a 2 and dividing by b 3 gives:
Since dV = dx dy dz is the volume for a rectangular differential volume element (because the volume of a rectangular prism is the product of its sides), we can interpret dV = ρ 2 sin φ dρ dφ dθ as the volume of the spherical differential volume element. Unlike rectangular differential volume element's volume, this differential volume ...