Mister rogers is fencing another new rectangular garden in his neighborhood. one side of the garden faces the road and needs to be pretty. the other three sides just need to be functional. the pretty fencing costs $35 per linear foot and the functional fencing costs $18 per linear foot. mr. rogers has $ 3000 to build his fence. what dimensions of the garden give him the maximum area?

length of the pretty side  and length of the side oppositte to the pretty side  =   37.91 ft

length of the other two sides  = 27.52 ft

Step-by-step explanation:

The mathematical problem is:

Max A = b1*h

subject to: 35*b1 + 18*(2*h + b2) <= 3000

Where

A: area of the garden

b1: length of the pretty side

b2: length of the side oppositte to the pretty side

h: length of the other two sides

Replacing with b1 = b2 and taking only the equality sign in the restriction (in the maximum all the money will be spent), we get:

35*b1 + 18*(2*h + b1) = 3000

35*b1 + 36*h + 18*b1 = 3000

53*b1 + 36*h = 3000

b1 = 3000/53 - (36/53)*h

Substituing in Area's formula  

A = (3000/53 - (36/53)*h)*h

A = (3000/53)*h - (36/53)*h^2

In the maximum, the derivative of A is equal to zero

dA/dh = 3000/53 - 2*(36/53)*h =

3000/53 - 72/35*h = 0

h = (3000/53)*(35/72)

h = 27.52 ft

then,

b1 = 3000/53 - (36/53)*27.52

b1 = 37.91 ft =b2

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.

his first public project he is asked to create a small scale drawing. the garden is a right triangle with base 25m and height 30m. are you trying to solve for the last side/hypotenuse? the irrational form of the answer would be 5\sqrt{61}