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adell [148]
3 years ago
7

Seorang pemilik toko sepatu ingin mengisi tokonya dengan sepatu laki-laki paling sedikit 100 pasang dan sepatu wanita paling sed

ikit 150 pasang. Seorang pemilik toko sepatu ingin mengisi tokonya dengan sepatu laki-laki paling sedikit 100 pasang dan sepatu wanita paling sedikit 150 pasang. Toko tersebut hanya dapat menampung 400 pasang sepatu. Keuntungan setiap pasang sepatu laki-laki adalah Rp 10.000,00 dan keuntungan setiap pasang sepatu wanita adalah Rp 5.000,00. Jika banyaknya sepatu laki-laki tidak boleh melebihi 150 pasang, maka tentukanlah keuntungan terbesar yang dapat diperoleh oleh pemilik toko.

Mathematics
1 answer:
taurus [48]3 years ago
5 0

Answer:

<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.

Step-by-step explanation:

We need to form a LPP( Linear Programming Problem) for the following problem and solve it to get the maximize the profit.

Let x denote the number of men's shoe and y denote the number of female shoes.

Maximize z= 10000x+5000y----------(1)

             x≥100 ( since a shoe store owner wants to fill his shop with at least    100 pairs of men's shoes )

            y≥150  (at least 150 pairs of women's shoes).

also x+y≤400   (The store can only accommodate 400 pairs of shoes).

         x≤150    (male shoes should not exceed 150 pairs).

⇒         100≤x≤150

Hence the optimal solution of an LPP always lie on the end points.

We have end points as:

(100,300), (150,250), (100,150),(150,150).

by putting these value in equation (1) we see which give the maximum solution.

for (100,300) i.e. x=100 and y=300:   z=2,500,000

for (150,250) i.e. x=150 and y=250:  z=2,750,000

for (100,150) i.e. x=100 and y=150: z=1,750,000

for (150,150) i.e. x=150 and y=150: z=2,250,000

Hence maximum profit is obtained at x=100 and y=300.

i.e. to maximize the profit:

<em>Number of men's shoes to be kept in store=100</em>

<em>and number of women's shoes to be kept in store=300</em>

<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.

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Hi there!

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xy + y² = 6

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(1)y + x(dy/dx)  + 2y(dy/dx) = 0

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Factor out dy/dx:

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Divide both sides by x + 2y:

dy/dx = -y/x + 2y

We need both x and y to find dy/dx, so plug in the given value of x into the original equation:

-1(y) + y² = 6

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At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.

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Why is 210 divided by 30 the same as 21 tens divided by 3 tens
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The length of a rectangle is 6 ft more than its width. The perimeter of the rectangle is 68 ft. What are the dimensions of the r
Anastaziya [24]
First off,we need to understandthe perimeter of a rectangle:
2(width + height)

We can then proceed to find the width. As it is 6 ft more than it's length,it can be written in:

w + 6
We can then find the equation of the perimeter:

2((2 + w) + w) = 68 \\ 2(2 + 2w) = 68 \\ 4 + 4w = 68 \\ 4(1 + w) = 68 \\ 1 + w =  \frac{68}{4}  \\ w = 17 - 1 \\ w = 16

When the width is 16ft, the length is:
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3 years ago
A rectangular cow pasture is enclosed on three sides by a fence and the fourth side is part of the side of a barn that is $400$
Rudiy27

Let us list out the given information:

Cost per foot = $5

Total fencing cost = $1200

Let L be the length of the barn that will maximize the area of the pasture and W be the width of the rectangular cow pasture.

From the given information, we need to do the fencing for three sides covering 1 L and 2 W's because other L will be the fourth side is part of the side of a barn. We need to write the equation based on the cost information.

If cost/foot is $5, then cost of length ' L ' will be 5L dollars and cost of '2W' width equals 10W dollars. Total cost is $1200. We can set up second equation as

5L + 10W = 1200. This can be simplified by dividing throughout the equation by 5 to get L+ 2W= 240. Solving for L we get L = 240 - 2W

Area of a rectangle is given by A = L ×W.

Substituting 240 - 2W for L we get the area function as..

A=(240 - 2W)×W= 240W-2W²

A = -2W²+240W.

When we have a function of the form f(x) = ax^2 + bx + c, then maximum value of the function can be obtained for x = -b/2a. Using that idea we can find the maximum area.

Here in the area function a = -2 and b = 240

For W = -b/2a = -240/2(-2) = -240/-4 = 60 feet, we get maximum area.

The question is asking us to find Length. We can substitute W = 60 in the equation L = 240 - 2W to find the Length.

L = 240 - 2(60) = 240 - 120 = 120 feet.

Conclusion: The length of the side parallel to the barn that will maximize the area of the pasture is 120 feet.

Note: Here the information 400 feet in the sentence "the fourth side is part of the side of a barn that is 400 feet long" given to distract us, we might not require that information to find the Length.

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3 years ago
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