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Question 1: The two shorter sides of a right triangle ABC are on the coordinate axes and its hypotenuse passes through the point (1,8): A (1,8) в We are interested in finding the vertices A and C such that the length of the hypotenuse is minimum. Lets us do it as follows: (a) Write the equation of a straight line whose slope is m and which passes through the point (1,8). (b) Find out the x-intercept and y-intercept of the line in (a) in terms of the slope m. (c) Let us denote the length of the hypotenuse AC as h. h is a function of the slope m. Show that m h(m) : + (8— т)2 (d) With our current knowledge of calculus, it is difficult to find m that minimizes h(m). So instead of minimizing h(m), let us minimize r(m) = h°(m). This will not change the location of the minimum m. To convince yourself graph both h(m) and r(m) = h°(m). Where do you locate the minimum? (e) Now show that 2. 64 16 + 65 m m - - 8 r(m) + (8— т)? — т? + 16т m m2

Question 1: The two shorter sides of a right triangle ABC are on the coordinate axes and its hypotenuse passes through the point (1,8): A (1,8) в We are interested in finding the vertices A and C such that the length of the hypotenuse is minimum. Lets us do it as follows: (a) Write the equation of a straight line whose slope is m and which passes through the point (1,8). (b) Find out the x-intercept and y-intercept of the line in (a) in terms of the slope m. (c) Let us denote the length of the hypotenuse AC as h. h is a function of the slope m. Show that m h(m) : + (8— т)2 (d) With our current knowledge of calculus, it is difficult to find m that minimizes h(m). So instead of minimizing h(m), let us minimize r(m) = h°(m). This will not change the location of the minimum m. To convince yourself graph both h(m) and r(m) = h°(m). Where do you locate the minimum? (e) Now show that 2. 64 16 + 65 m m - - 8 r(m) + (8— т)? — т? + 16т m m2

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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(f) Use your knowledge of first derivatives to find the minimum value of r(m) and the corresponding m that achieves it. In particular show that the minimization amounts to solving the following equation: m4 – 8m3 + 8m – 64 = 0 - (g) Solve the polynomial equation in (e) and substitute the appropriate value of m in h(m) to find the length of the hypotenuse. Hence find the coordinates of the vertices A and C.
Transcribed Image Text:(f) Use your knowledge of first derivatives to find the minimum value of r(m) and the corresponding m that achieves it. In particular show that the minimization amounts to solving the following equation: m4 – 8m3 + 8m – 64 = 0 - (g) Solve the polynomial equation in (e) and substitute the appropriate value of m in h(m) to find the length of the hypotenuse. Hence find the coordinates of the vertices A and C.
Question 1: The two shorter sides of a right triangle ABC are on the coordinate axes and its hypotenuse passes through the point (1,8): y A (1,8) в We are interested in finding the vertices A and C such that the length of the hypotenuse is minimum. Lets us do it as follows: (a) Write the equation of a straight line whose slope is m and which passes through the point (1,8). (b) Find out the x-intercept and y-intercept of the line in (a) in terms of the slope m. (c) Let us denote the length of the hypotenuse AC as h. h is a function of the slope m. Show that: 2 h(m) - 8 + (8— т)2 (d) With our current knowledge of calculus, it is difficult to find m that minimizes h(m). So instead of minimizing h(m), let us minimize r(m) = h°(m). This will not change the location of the minimum m. To convince yourself graph both h(m) and r(m) = h°(m). Where do you locate the minimum? (e) Now show that 64 + (8— т)? — т? + m2 m - 16 r(m) + 65 m 16т
Transcribed Image Text:Question 1: The two shorter sides of a right triangle ABC are on the coordinate axes and its hypotenuse passes through the point (1,8): y A (1,8) в We are interested in finding the vertices A and C such that the length of the hypotenuse is minimum. Lets us do it as follows: (a) Write the equation of a straight line whose slope is m and which passes through the point (1,8). (b) Find out the x-intercept and y-intercept of the line in (a) in terms of the slope m. (c) Let us denote the length of the hypotenuse AC as h. h is a function of the slope m. Show that: 2 h(m) - 8 + (8— т)2 (d) With our current knowledge of calculus, it is difficult to find m that minimizes h(m). So instead of minimizing h(m), let us minimize r(m) = h°(m). This will not change the location of the minimum m. To convince yourself graph both h(m) and r(m) = h°(m). Where do you locate the minimum? (e) Now show that 64 + (8— т)? — т? + m2 m - 16 r(m) + 65 m 16т
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