Optimization Of Exponential And Logarithmic Functions

9.12. Given v - 4ic", (a) find Ihc critical values and (6) determine whether the function is maximized or minimized.

a) By the product rule.

Since there i\ no value of i for which » 0. or far which r* - tl,

b) f « 4ru(3) ♦ (3« t- IKI2f") - I2r"<3« ♦ 2)

At x - I2e 'OV And / 12(036788) > 0. The (unction b minimized

9.13. Redo Problem 9.12, given y «= Sir a) y' - 5*< -0.2e y*(S) - 0

b) f - 5r **(-Q2) 02»H -If ^«r "^(02« - 2) AtJf*5,y«r '(1 2) And y (0.36788K-1><0 Ihc function t\ at a maximum

9.14. Redo Problem 9.12. given v * ln(r - Kt f 20).

Multiplying both odes by t■ - 8i * 20 goes 2« - K 0 and i « 4.

At ( • jy » &>0 Ihe function is at a minimum.

9.15. Redo Problem 9.12. given v - In (2r - 20* 4- 5).

(2r - 20* ♦ 5H4) - (4* 20)(4r - 20) ' ' (2r - 20* 4 5):

At i ■ 5. f 3 -1WV2025<0 The function naia maximum.

9.16. Given the (unction r ln(2r; 12r ♦ v lOv), (a) tind the critical values and (b) indicate whether the function is at a maximum or minimum.

. (2^ - 12* ♦ y* - 10yM4) - (4* - 12X4* - 12) ' " (2i* —12» 4- y1 - Ifly)1

(2*^ - 12* 4 r - KH H2) - (2y - I0H2> - 10) (2x: 12« ♦ y> 10y?

At « - 3. V « 5. * 0 ■ ;„ With :„. z„< 0 and rMir, >(:„)'. the function b .n a maximum

9.17. Redo Problem 9.16. given : - In (f - 4i 4 V - by).

. ,(^-4*4V-6y)(2)-(2x-4K2*-4) ' " (TJ-4,4V-ft>f

_ U' - 4« 4 3y* - 6yM6) - (6v - 6)(6y - 6) (r* - 4* 4 3y* - 6y?

_ -(2c-4M6y-6) ' (** - 4* 4 3y* - 6y)J " ^ At i 2. ji I.?.. 0 - :„. With r„, :„< 0 and c„r„ >(z,%):. the function i\ at a maximum

b) l Ifcing (he product rule,

Ibcn tcNiinü the cttm paitialv

^ - (ta - 6X2y - gj^ W-*» , Z t At i I. $ ~ 4. r„ - 0 - Ihc function u at a mínimum \ince j„ >fl and >(r„)í.

Solving (910) and (9.//) simultaneously. * • H. y - 10

6) rM « (4t - 12 - 2yM4« - 12 - 2y)ru*1 * ^ >.»•

.V - (-1» ♦ 2y - 4M - 2t ♦ 2y- 4)«^ + «fl^- »*- W

• o ♦ 4r m >0 z„*0+2r m>0 toting tbv mixed partiaK by the product rule.

Evaluated at i « 8. y - 10. r„ »0-2- ;„. Since >0 and Ct.^„>(:„),. the function ts at a mínimum

9.20. (¡iven the demand function

(a) Determine the quantity and price at which total revenue will be maximized and (h) test the second-order condition,

Bv the product rule.

Since (8.25r a 0 for any value of Q.\ - 0.02C> - 0. <> - 50

Substituting Q - 50 in (9/2). P - 82V i-"*» » 825r». And P - 8.25(0.36788) - 3.04. /») By the produci rule.

- (K25r •"CVM 0.02) Ml - 0.02(?M-002)(8.25e aJKv) - (-0.02M8.2.V ♦WVX2 - 0020)

Evaluated at 0 * 50. ^TR/rfff* =» (-0 02X8.25* 'HO " -0-165(036788)<0. TK >% at a maw mum

9.21. (u) Find the price and quantit) lhal will maximize total revenue, given the demand function P - I2.50r ucn*v. (b) Cheek the second-order condition.

« (I2.50r I - 0005C» - 0 I 0.005(2 - 0 Q — 200 Thus, P - 12.50* - I2.50r ' - 12.50(0.36788) - 4.60

b) » (I2.50r 0.005) ♦ (I - 0.005^K- 0.005HIZSOr JQ-

Evaluated at (j = 200. ifTRJdQ* • (-0 005X12.50* 'Xl> -0.0625(0 36788)<0 Ihe function is maximized.

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