Find the point on the graph of the function f(x) = (x+1)^2 that is closest to th

Question

Find the point on the graph of the function f(x) = (x+1)^2 that is closest to the point (-5, 1). Round all numerical values in your answer to four decimal places. (3.3811, 1.9370) (-3.3811, 2.3918) (-2.3918, 1.9370) (-1.9370, 3.3811) (2.3918, 1.9370) Find the length and width of a rectangle that has an area of 128 square feet and whose perimeter is a minimum. l = 64 feet; w = 2 feet l = 16feet; w = 8 feet l = 64 squareroot 2 feet; w = squareroot 2 feet l = w= 8 squareroot 2 feet l = 32 feet; w=4feet On a given day, the flow rate F (cars per hour) on a congested roadway is given by F = v/28 + 0.07v^2, where v is the speed of the traffic in miles per hour. What speed will maximize the flow rate on the road? Round your answer to the nearest mile per hour. 20 miles per hour 18 miles per hour 19 miles per hour 17 miles per hour 10 miles per hour

Explanation / Answer

7) consider a point on the curve ( x, (x+1)^2) now distance between given point

d^2 = (x+5)^2 +((x+1)^2 -1)^2

differentiate w.r.t x and make = 0

2*(x+5) + 2*((x+1)^2-1) * 2*(x+1) = 0

x+5 + 2(x+1) (x^2+2x)= 0

solving , we have x = -2.3918 .

so answer is OPTION C -----ANSWER 7

8)

A = L * B = 128

perimeter = 2L+2B

apply lagrange theorem ,

f = 2L+2B - lambda (LB-128)

df/dL = 2- L*lambda => lambda = 2/L

df/dB = 2-lambda (B) => lambda = 2/B

=> 2/L = 2/B => L= B

put in LB = 128 , we have L =B = sqrt(128) = 8*sqrt(2) -----OPTION D

9)

diffrentiate F wrt v

dF/dv = (28+0.07v^2) *1 -v*(0.14 v) /(28+0.07v^2)

dF/dv= 0 => 28 +0.07v^2 -0.14v^2 = 0

solving we get v = +- 20 , as speed cannot be -ve , we have +20 as answer so , OPTION A

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