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oee [108]
3 years ago
10

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?
Mathematics
1 answer:
Oksanka [162]3 years ago
3 0

Answer:

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

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Answer:

Width=13cm

Length=18cm

Step-by-step explanation:

Width of the Initial Photograph=9cm

Length of the Initial Photograph=14cm

If width and length are increased by an equal amount, say x

New Width=9+x

New Length=14+x

Area of the new photo is 108 square centimeters greater than the area of the original photo.

Area of Original Photo=14*9=126cm²

Aeea of the New Photo=126+108=234cm²

Therefore:

(9+x)(14+x)=234

126+9x+14x+x²-234=0

x²+23x-108=0

x²+27x-4x-108=0

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(x+27)(x-4)=0

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x=-27 or x=4

Since x≠-27

x=4cm

The dimensions of the new photo are:

Width=9+x=9+4=13cm

Length=14+x=14+4=18cm

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2 years ago
If log(16)=s, what is the value of 2/3 log(6400) - 1/3 log(1600) in terms of s?
eduard

The value of   \dfrac{ 2}{3} log(6400) - \dfrac{1}{3} log(1600)  in terms of s will be

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Here we have the expression:-

\dfrac{ 2}{3} log(6400) - \dfrac{1}{3} log(1600)

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Answer:

All of the above

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