# 1.) f (x) = 3x³- 3x² – 3x + 4

Get the firstderivative and equate to zero as follows:

9x²- 6x -3 = 0

Simplify theequation to get

3x² – 2x – 1=0

3x² – 3x + x-1=0

3x (x-1) +1 (x-1)=0

(3x+1) (x -1)=0

The criticalpoints are x= 1 and x= -.

However, theinterval provided is [-1, 0].

Therefore, theabsolute maximum value is 4 at x=0.

Also, theabsolute minimum value is 1 at x= -.

2.) f (x) = 2 -8x

Get the firstderivative and equate to zero

0-8=0

-8=0

This shows thatthe equation has no global maxima and minima.

The intervalgiven is [-4, 3].

The absolutemaximum value is 2 at x=0.

The absoluteminimum value is 0 at x=0.

3.) P(x) = R(x) –C(x)

a.) P(x) = -x² + 600x – 6000

Find derivativeof P(x) as follows:

-x + 600 = 0

b.) The profit ismaximized for a production of 600 units

4.) Q = 6xy² andx + y² = 12

Substitute for y²= 12-x

Q = 6x (12 – x)

Q = 72x – 6x²

Get the firstderivative and equate to zero as follows:

72 – 12x = 0

Therefore, X = 6and Y = √6.

5.) Let x by thewidth of each area, and y the length.

6x + 4y = 1008yards (yd)

Area = 3xy

Substitute for y= 252 – 1.5x

Therefore, area =3x (252 – 1.5x).

Area =756x -4.5x²

Get thederivative and equate to zero as follows:

756 – 9x = 0

Therefore, X= 84.

Substitute for Xso as to obtain Y as follows:

Y= 252 -1.5 (84)

Therefore, Y=126.

B. The length ofthe shortest side is 126 yards and the length of the longest side is252 yards.

The largest totalarea that can be enclosed is 31,752 yd².

6.) Let x be thewidth and y be the length of the rectangle

2x + 2y = 53

Area = xy

Substitute for y= 26.5 – x as follows:

Area= x (26.5 -x)

Get the firstderivative and equate to zero as follows:

26.5 – 2x = 0

Therefore, X=13.25.

Substitute for Xto obtain Y as follows:

Y = 26.5 – 13.25

Therefore, Y =13.25.

A. The length ofeach side is the same and measures 13.25 ft.

The area is175.5625 ft².

7.) P(x) = R(x) –C(x)

P(x) = (9x – 2x²)- (x³ – 4x² + 3x + 1)

P(x) = 9x – 2x²- x³ + 4x² – 3x – 1

P(x) = -x³ + 2x²+ 6x – 1

Get thederivative and equate to zero as follows:

-3x² + 4x + 6 =0

3x² – 4x – 6 =0, where a= 3, b=-4, and c=-6

X=

X= 10.0474 sincewe can only take the positive value of units.

Therefore, theproduction level for the maximum profit is about 10 units.

But, P(x) = -x³+ 2x² + 6x – 1.

Substitute for xand get Profit as follows:

Profit =-753.1278.

Therefore, theprofit is about -\$753, which translates to a loss of \$753.

8.) (5x² – 4y²) = (13)

(5x²) – 8y= 0

10x – 8y = 0

Therefore, =.

The slope at (√5,√3) =.

Therefore, theslope =.

9.) (x²y-4x²-8) = (0)

(x²y) – (4x²)– (8)= 0

y (x²)+ x² (y)- 8x = 0

2xy + x² – 8x = 0

Therefore,

The slope at (2,6) =.

The slope at (2,6) =.

Therefore, theslope = -2.

10.) (6xy -7x + y) = (-6)

(6xy) – (7x)+ (y)= 0

6xy(6x)- 7 +=0

6x+ 6y -7 +=0

Group the together to obtain:

(6x+1)= 7-6y

Therefore, =.

The slope at(1,)=.

The slope at(1,)=

Therefore, theslope =.