Thus, the point we have found is a local minimum. 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. Solving for s, we get that s = (4*350) 1/3 = 2 * 5 2/3 * 7 1/3. Find (a) the volume and (b) the total surface area of a right-angled triangular prism of length 80cm and whose triangular end has a base of 12cm and. 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. Example: What is the volume of a prism where the base area is 25 m 2 and which is 12 m long: Volume Area × Length. ≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ □ □ □ ⊲ ⊳ The Total Surface Area of Triangular Prism formula is defined as the total amount of two-dimensional space occupied by all the faces of the Triangular Prism and is represented as TSA (h (S a + S b + S c))+(1/2 sqrt ((S a + S b + S c)(S a + S b-S c)(S b + S c-S a)(S c + S a-S b))) or Total Surface Area of Triangular Prism (Height of. Re-read your post before hitting submit, does it still make sense.Show your work! Detail what you have tried and what isn't working. Use proper spelling, grammar and punctuation.Give context and details to your question, not just the equation.Help others, help you! How to ask a good question Asking for solutions without any effort on your part, is not okay. Beginner questions and asking for help with homework is okay. Post your question and outline the steps you've taken to solve the problem on your own.
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