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April 23, 2014

April 23, 2014

Posted by **Hannah** on Saturday, April 30, 2011 at 7:58pm.

e^x=0 and then (x+2)(x-2) so x=2, -2

2) differentiate: y=ln(6x^2 - 3x + 1)

1/(6x^2 - 3x + 1) * 12x-3

3) differentiate: y=e^-3x+2

-3 * e^-3x+2

4) evaluate: 2^4-x=8

2^4-x = 2^3

4-x = 3

-x=-1 so x=1

5) differentiate: x^3 + y^3 -6 =0

3x^2 + 3y^2

6) A rectangular garden has an area of 100 square meters for which the amount of fencing needed to surround the garden should be as small as possible.

a) draw a picture of a rectangle and select appropriate letters for the dimensions

I chose x and y

b)determine the objective and constraint equations

objective: A=xy

constraint= 100=xy

c) find the optimal values for the dimensions.

I am not sure for this one

Thank you for your help!!!

- Math(Please check. Thank You) -
**bobpursley**, Saturday, April 30, 2011 at 8:28pm1. e^x can never equal zero, but can approach it as x>>-infinity

But the roots are in fact -2,2

2 correct

3 e^(-3x+2)=e^-3x * e^2

then y'= -3e^-3x * e^2=-3e^(-3x+2)

4, correct

5. No.

x^3+y^3=0 You can do implicit differentation, but I am not certain you know that.

y^3=x^3

y= cubroot x^3=x

y'=1

6) xy=100 constraint

Perimeter= 2x+2y objective

dP/dx= 2+ d2(200/x)/dx= 2-400/x^2 =0

2x^2=400

x=10 Y=10 is minimum fencing

- Math(Please check. Thank You) -
**Mgraph**, Saturday, April 30, 2011 at 8:50pm5) (x^3+y^3-6)'=3x^2+3y^2*y'=0

y'=-x^2/y^2, where y=(6-x^3)^(1/3)

- Math(Please check. Thank You) -
**bobpursley**, Saturday, April 30, 2011 at 8:53pmI agree, Mgraph, thanks.

- Math(Please check. Thank You) -
**Mgraph**, Saturday, April 30, 2011 at 8:53pm5) (x^3+y^3-6)'=3x^2+3y^2*y'=0

y'=-x^2/y^2, where y=(6-x^3)^(1/3)

- Math(Please check. Thank You) -
**Hannah**, Saturday, April 30, 2011 at 11:22pmThank you!!

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