![]() ![]() ![]() Their studies on triangles, parallelograms, and three-dimensional shapes have greatly influenced contemporary understanding of geometry and the surface area of various shapes, including triangular prisms. While there is no definitive historical account of the origin of the triangular prism or its surface area concept, it can be traced back to ancient Greece, where mathematicians like Euclid and Pythagoras laid the groundwork for modern geometry. Moreover, artists and designers frequently employ triangular prisms in their creations, making the knowledge of surface area invaluable for conceptualizing and executing their work. In packaging design, calculating the surface area of a triangular prism helps optimize material usage, reduce waste, and minimize costs. For instance, in construction and architecture, the surface area plays a role in determining the stability and strength of structures, as well as insulation and energy efficiency. Triangular prisms, like other three-dimensional shapes, have numerous real-life applications that make understanding their surface area essential. The concept of surface area has broad applications in various fields, including engineering, architecture, and design, where it is crucial to estimate material requirements, costs, and structural integrity. A triangular prism consists of two congruent triangles at the ends, known as bases, connected by three parallelogram-shaped lateral faces. Thus, the point we have found is a local minimum.The surface area of a triangular prism is a key concept in geometry that pertains to the total area covering the external faces of the three-dimensional shape. ![]() The second derivative of this guy is strictly positive for positive s, implying the function is concave up for positive s. To do so you must take the second derivative. ![]() We'll end up with h = 2 * 5 2/3 *7 1/3 / sqrt(3).ĮDIT: It's a bit pedantic, but technically you have to make sure that it's a local minimum at the value of s that I've found. From there, we can easily find the height by substituting into our previous formula. We want to find the minimum so we set SA' = 0. SA = 2(sqrt(3)/4)s 2 + 3sh (the first term is the 2 triangular parts and the second term is the three lateral, rectangular parts).Īs a function of s alone, we have SA = 2(sqrt(3)/4)s 2 + 4sqrt(3)350/s. This is equivalent to h = 4*350/(sqrt(3)s 2 ). V = (sqrt(3)/4)hs 2 = 350 cm 3 (I converted mL to cm 3 for ease). Then the area of the base is (sqrt(3)/4)s 2. Let s be the base of the triangle and h be the height. This is an ordinary optimization problem so it requires the use of basic calculus.
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