Tiêu đề 71 Maxima Minima.pdf ✅

Here, h and x are constants, while θ is a variable. Differentiating with respect to θ, L ′ = −h csc θ cot θ + x sec θ tan θ Setting L ′ = 0, 0 = −h csc θ cot θ + x sec θ tan θ sec θ tan θ csc θ cot θ = h x tan3 θ = h x tan θ = √ h x 3 From the Quotient and Pythagorean identities, csc θ = (x 2 3 + h 2 3) 1 2 h 1 3 , sec θ = (x 2 3 + h 2 3) 1 2 x 1 3 Substitute the values to solve for L, L = h [ (x 2 3 + h 2 3) 1 2 h 1 3 ] + x [ (x 2 3 + h 2 3) 1 2 x 1 3 ] L = (x 2 3 + h 2 3) 3 2 Therefore, the relationship between the three distances is L 2 3 = x 2 3 + h 2 3
2.2. The Longest line that passes through two perpendicular corridors [DERIVATION] From the figure, the length of L, in terms of the widths of the corridors and the angle of inclination of the line, is L = a csc θ + b sec θ Which results in the same procedure as the previous formula. Hence, the relation is L 2 3 = a 2 3 + b 2 3 2.3. The maximum area of a rectangle with a known perimeter Let the dimensions of the rectangle be x and y. [DERIVATION] The perimeter of the rectangle is P = 2x + 2y Solve for y: y = P 2 − x From the formula of the area of a rectangle, A = xy A = x ( P 2 − x) A = P 2 x − x 2 Differentiating the formula with respect to x, A ′ = P 2 − 2x
Setting A ′ = 0, it results to x = P 4 Then solving for y gives y = P 4 . Since x = y = P 4 , then the rectangle should be a square. 2.4. Minimum perimeter of a rectangle with a known area Let the dimensions be x and y. [DERIVATION] The area is A = xy Solve for y: y = A x The perimeter of the rectangle is P = 2x + 2y Substituting the values, P = 2x + 2 ( A x ) P = 2x + 2A x From differentiation and setting A ′ = 0, it results to x = √A Therefore, it also results in y = √A. Since x = y = √A, then the rectangle is also a square.