Optimisation, Derivatives, Integrals and Matrices

Optimisation, Derivatives, Integrals and Matrices 2021 1 Maxima and Minima Question 1: Optimize the following function by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum and (3) the values of the relative extrema for the function. 𝑓 (𝑥) = 𝑥 2 + 6𝑥 + 9 A) 𝑥 = −3, relative maximum, 𝑓 (−3) = 0 B) 𝑥 = 3, relative minimum, 𝑓 (−3) = 0 C) 𝑥 = −3, relative minimum, 𝑓 (−3) = 0 D) 𝑥 = −3, relative minimum, 𝑓 (−3) = −1 E) None of the above Answer: C) Find critical point by taking the first derivative and setting it equal to zero: 𝑓 0 (𝑥) = 0; 2𝑥 + 6 = 0 which solves for 𝑥 = −3. Find the second derivative 𝑓 00 (𝑥) = 2 > 0 meaning that the function is convex (U-shaped, positive 2nd derivative → concavity faces upwards), hence at 𝑥 = −3 we have a relative minimum. 𝑓 (−3) = −32 −6×3+9 = 0 Question 2: Optimize the following function by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum and (3) the values of the relative extrema for the function. 𝑓 (𝑥) = 𝑥 3 + 6𝑥 2 − 96𝑥 + 23 A) 𝑥 = 4, 𝑓 ( 4) = −201 relative minimum; 𝑥 = −8, 𝑓 (−8) = 663 relative maximum. B) 𝑥 = −4, 𝑓 (−4) = 200 relative minimum; 𝑥 = 8, 𝑓 ( 8) = −665 relative maximum. C) 𝑥 = 4, 𝑓 ( 4) = −201 relative minimum; 𝑥 = 8, 𝑓 ( 8) = 663 relative maximum. D) 𝑥 = 4, 𝑓 ( 4) = −154 relative minimum; 𝑥 = −8, 𝑓 (−8) = 653 relative maximum. E) None of the above 1 Answer: A) Find critical points by taking the first derivative and setting it equal to zero: 𝑓 0 (𝑥) = 0; 3𝑥 2 + 12𝑥 − 96 = 0 which means 3 (𝑥 2 + 4𝑥 − 32) = 0; (𝑥 − 4) (𝑥 + 8) = 0 so critical values are 𝑥 = 4 and 𝑥 = −8. Check the sign of the second derivative 𝑓 00 (𝑥) = 6𝑥 + 12 which is positive for 𝑥 = 4 ( 𝑓 00 ( 4) = 24 + 12 = 36 > 0 relative minimum) and negative for 𝑥 = −8 ( 𝑓 00 (−8) = −48 + 12 = −36 < 0 relative maximum). By substituting each x-value in 𝑓 (𝑥) we’ve got 𝑓 ( 4) = −201 and 𝑓 (−8) = 663 Question 3: A department store is fencing off part of the store for children to meet and be photographed with Santa Claus. They have decided to fence off a rectangular region of fixed area 800 𝑓 𝑡 2 . Fire regulations require that there be three gaps in the fencing: 6 ft openings on the two facing sides and a 10 ft opening on the remaining wall (the fourth side of the rectangle will be against the building wall). Find the dimensions that will minimize the length of fencing used Figure 1: Picture of the store Answer: Let 𝑥 be the side with 10 ft gap and 𝑥 be the side 6 ft gap. It is given that 𝑥𝑦 = 800 and we want to minimize 𝑓 = (𝑥 − 10) + 2 (𝑦 − 6) Which is a constrained optimisation problem in two variables. However, since we know that the area has to be 𝑥𝑦 = 800, we can re-write 𝑦 = 800 𝑥 and substitute this in the objective function, i.e. the function that we want to optimise. 𝑓 = (𝑥 − 10) + 2 ( 800 − 6) 𝑥 and the domain of this function is [10, ∞] because we need at least 10 ft gap. To find minimal value 𝑓0 =1− 1600 𝑥2 𝑥 = 40 and 𝑦 = 20 2 =0 2 Derivatives √ Question 4: Let 𝑓 (𝑥) = 1 + 𝑥 2 . Compute 𝑓 0 (𝑥) , 𝑓 00 (𝑥) and determine whether the function is concave, convex or neither. Answer: 𝑓 0 (𝑥) = √𝑥 1+𝑥 2 3 and since 𝑓 00 (𝑥) = ( 1 + 𝑥 2 ) − 2 > 0, then 𝑓 (𝑥) is convex. Question 5: (This question is not directly related with the material, but it is very easy to solve if you understand derivative notion.) Calculate the partial derivatives 𝑓𝑦 (𝑥, 𝑦) , 𝑓𝑥 (𝑥, 𝑦), 𝑓𝑦𝑦 (𝑥, 𝑦) , 𝑓𝑥𝑥 (𝑥, 𝑦) and 𝑓𝑦𝑥 (𝑥, 𝑦) for the following functions. (i) 𝑓 (𝑥, 𝑦) = 𝑥 3 + 3𝑥𝑦 2 + 2𝑦 (ii) 𝑓 (𝑥, 𝑦) = ( 3𝑥 2 + 2𝑦) 2 Answer: i) By calculating 𝑓𝑥 = 𝜕𝑓 𝜕𝑥 you want to find out how the function changes as only 𝑥 changes marginally, whilst 𝑦 is kept constant. This means that you differentiate the function with respect to (wrt) 𝑥 , treating 𝑦 as this was any constant. So in our case 𝑓𝑥 = 𝜕𝑓 𝜕𝑥 = 3𝑥 2 + 3𝑦 2 . Notice that the last term 2𝑦 has disappeared. Now the cross- partial derivative. We take the derivative wrt 𝑥 which we just found, differentiate wrt 𝑦 and find 𝑓𝑦𝑥 = Now we differentiate the initial function wrt to 𝑥 twice, 𝑓𝑥𝑥 = 𝑓𝑦 = 2 𝜕 𝑓 𝜕𝑥 2 𝜕𝑓 𝜕𝑦 𝜕2 𝑓 𝜕𝑥 𝜕𝑦 = 6𝑦 𝑓𝑦𝑦 = 𝜕2 𝑓 𝜕𝑦 2 = 6𝑥 Now ii) 𝜕𝑓 𝜕𝑥 = 12𝑥 ( 3𝑥 2 + 2𝑦) = 36𝑥 3 + 24𝑥𝑦 𝑓𝑥 = 𝑓𝑦𝑥 𝜕2 𝑓 = 𝜕𝑦𝜕𝑥 = 24𝑥 𝑓𝑥𝑥 = 𝜕2 𝑓 𝜕𝑦 2 = 108𝑥 2 + 24𝑦 3 = 6𝑦 . = 6𝑥 . And now repeat all the above but wrt to 𝑦 . = 6𝑥𝑦 + 2 𝑓𝑥 𝑦 = 𝜕2 𝑓 𝜕𝑦𝜕𝑥 𝑓𝑦 = 𝜕𝑓 𝜕𝑦 = 4 ( 3𝑥 2 + 2𝑦) = 12𝑥 2 + 8𝑦 𝑓𝑥 𝑦 = 𝜕2 𝑓 𝜕𝑥 𝜕𝑦 = 24𝑥 𝑓𝑦𝑦 = 𝜕2 𝑓 𝜕𝑦 2 =8 3 Integrals Question 6: Solve the following integrals ∫ ( 3𝑥 2 + 6𝑥 + 7 + 𝑥1 ) 𝑑𝑥 ∫ √ (ii) ( 4 3 𝑥 − 𝑥2 + 𝑥1 )𝑑𝑥 ∫ √ 3𝑥 + 4 𝑑𝑥 (iii) ∫ (iv) 2𝑥 2𝑒 2𝑥 𝑑𝑥 ∫1 2 (v) 0 [ 4𝑥 ( 13 𝑥 3 + 1) + 2𝑥 3 ] 𝑑𝑥 (i) Answer: i) is simply the sum of the integral of each term. Basic rule of integration ∫ 𝑥 𝑘 𝑑𝑥 = 𝑥 𝑘+1 𝑘+1 + 𝑐 and ∫ 𝑘𝑑𝑥 = 𝑘𝑥 + 𝑐 ∫ 2 1 ∫ ∫ 2 ( 3𝑥 + 6𝑥 + 7 + ) 𝑑𝑥 = 𝑥 ∫ 3𝑥 𝑑𝑥 + 6𝑥 𝑑𝑥 + ∫ 7 𝑑𝑥 + 1 𝑑𝑥 𝑥 ∫ ∫ 2 𝑥 𝑑𝑥 + 6 =3 ∫ 𝑥 𝑑𝑥 + ∫ 7 𝑑𝑥 + 1 𝑑𝑥 𝑥 1 𝑥 2+1 + 6 1 𝑥 1+1 + 7𝑥 + 𝑙𝑛(𝑥) + 𝑐 2+1 1+1 1 1 = 3 𝑥 3 + 6 𝑥 2 + 7𝑥 + 𝑙𝑛(𝑥) + 𝑐 3 2 3 2 = 𝑥 + 3𝑥 + 7𝑥 + 𝑙𝑛(𝑥) + 𝑐 =3 ii) The trick is to re-arrange the function in a way that makes integration more convenient. So ∫ 1 ( 4𝑥 3 − 12 𝑥 + 𝑥1 )𝑑𝑥 4 ∫ √ ( 4 3 𝑥 − 𝑥2 + 𝑥1 )𝑑𝑥 = ∫ 1 1 1 𝑥 3 +1 1 1 2 − ( 4𝑥 − 𝑥 + )𝑑𝑥 = 4 1 𝑥 + 𝑙𝑛(𝑥) + 𝑐 2 𝑥 +1 22 3 1 3 4 =4 𝑥3 4 3 1 4 − 𝑥 2 + 𝑙𝑛(𝑥) + 𝑐 1 4 4 = 3𝑥 3 − 𝑥 2 + 𝑙𝑛(𝑥) + 𝑐 iii) use the method of 𝑢 -substitution. So if 𝑢 = 3𝑥 + 4 then intergal as ∫ 1√ 𝑢 3 𝑑𝑢 or ∫ 1 12 𝑢 3 = 3; and 𝑑𝑥 = 31 𝑑𝑢 . Thus, re-write the above 𝑑𝑢 𝑑𝑥 𝑑𝑢 ∫ 1 1 1 𝑢 2 𝑑𝑢 = 3 3 1 = 3 ∫ 1 𝑢 2 𝑑𝑢 1 1 2 1 𝑢 2 +1 +1 12 3 𝑢2 33 3 2 = ( 3𝑥 + 4) 2 9 = iv) For the second one use integration by parts. Remember ∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢 . A rule of thumb for the choice of 𝑢 is LIATE: Choose the u to be the first function that comes first in this list: L: Logarithmic Functions I: Inverse Trigonometric A: Algebric T: Trigonometric E: Exponential Let 𝑢 = 2𝑥 2 and 𝑑𝑣 = 𝑒 2𝑥 then solve ∫ 2 2𝑥 2𝑥 𝑒 𝑑𝑥 = 2𝑥 2 1 2𝑥 2 2 2𝑥 =𝑥 𝑒 The last term 2 ∫ 𝑒 ∫ − ∫ −2 1 4𝑥 𝑒 2𝑥 𝑑𝑥 2 𝑥𝑒 2𝑥 (1) 𝑑𝑥 𝑥𝑒 2𝑥 𝑑𝑥 requires an extra step in terms of integration by parts. We need to pick 𝑢 and 𝑑𝑣 once 5 again. So let Let 𝑢 = 𝑥 and 𝑑𝑣 = 𝑒 2𝑥 ∫ 𝑥𝑒 2𝑥 ∫ 1 1 𝑑𝑥 = 𝑥 𝑒 2𝑥 − 1 𝑒 2𝑥 𝑑𝑥 2 2 1 1 = 𝑥 𝑒 2𝑥 − 𝑒 2𝑥 + 𝑐 2 4 1 1 2𝑥 = 𝑒 (𝑥 − ) + 𝑐 2 2 Now we can put this back in Equation 1 ∫ 5 1 1 2𝑥 2𝑒 2𝑥 𝑑𝑥 = 𝑥 2𝑒 2𝑥 − 2 ( 𝑒 2𝑥 (𝑥 − )) + 𝑐 2 2 1 = 𝑥 2𝑒 2𝑥 − 𝑒 2𝑥 (𝑥 − ) + 𝑐 2 1 2𝑥 2 = 𝑒 (𝑥 − 𝑥 + ) + 𝑐 2 4 4 5 𝑥 + 56 𝑥 3 + 2𝑥 2 ] 10 = [ 15 × 15 + iv) [ 15 6 5 5 4 × 1 3 + 2 × 12 ] − [ 15 0 + 65 0 + 2 × 0] = 3.466 6 4 Matrices Question 7: Consider the following matrices 1 2 3 0 ! 1 ª © 𝐵 = ­­1 3 ®® 𝐴= −1 3 0 «4 −1¬ (i) calculate 𝐴 × 𝐵 and 𝐵 × 𝐴 (ii) find the determinant of 𝐴 × 𝐵 (iii) find the inverse of 𝐴 × 𝐵 Answer: i) 𝐴 × 𝐵 will “inherit” the number of rows of 𝐴 and the number of columns of 𝐵 . Thus, is going to be a 2 × 2 matrix. We do a “row-by-column” product in which the elements of the first row of 𝐴 are multiplied to the elements of the first column of 𝐵 , then first row of 𝐴 and second column of 𝐵 and so forth ! 0 1 ! ! ª 2 3 ©­ 1 × 0 + 2 × 1 + 3 × 4 1 × 1 + 2 × 3 + 3 (− 1 ) 14 4 𝐴×𝐵 = = 1 3 ®® = −1 3 0 ­ −1 × 0 + 3 × 1 + 0 × 4 −1 × 1 + 3 × 3 + 0 × − 1 3 8 «4 −1¬ 1 Similarly for 𝐵 × 𝐴, but this will “inherit” the number of rows of 𝐵 and columns of 𝐴, thus will be a 3 × 3 matrix. −1 3 0 © ª 𝐵 × 𝐴 = ­­ −2 11 3 ®® 5 12¬ « 5 ii) Determinant of 𝐴𝐵 . Given a generic matrix: 𝑎 𝑏 𝐴= 𝑐 𝑑 ! The determinant 𝑑𝑒𝑡 (𝐴) = 𝑎 × 𝑑 − 𝑏 × 𝑐 which in our case is 14 × 8 − 3 × 4 = 100 iii) The inverse of a matrix. Given the generic matrix stated above, the inverse is 𝐴 −1 𝑑 −𝑏 = 𝑑𝑒𝑡 (𝐴) −𝑐 𝑎 1 ! Hence 𝐴𝐵 −1 8 −4 1 = 100 −3 14 7 ! 𝐴𝐵 −1 = 8 100 3 − 100 4 − 100 ! 14 100 It is easy to verify that this is correct. The relationship between a matrix and its inverse is 𝐴𝐴−1 = 𝐼 where 𝐼 is the identity matrix, in this case 𝐼= 1 0 0 1 8 !