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3 years ago

Solve the following problems manually or using the MS Excel

Mathematics

1 answer:

ozzi3 years ago

7 0

Answer:

(a) in the step-by-step explanation

(b) The optimal solution is 8 chairs and 4 tables.

Step-by-step explanation:

(a)

C: number of small chairs

T: number of large tables

Maximize Income = 9T + 3C

Restrictions:

Wood: 4T+C<=24

Glazing: 2T+C<=16

In the graph its painted in green the "feasible region" where lies every solutions that fit the restrictions.

One of the three points marked in the graph is the optimal solution.

Point 1 (C= 16, T= 0)

Income = 9*0+3*16=$ 48

Point 2 (C=8, T=4)

Income = 9*4+3*8 = $ 60

Point 3 (C=0, T=6)

Income = 9*6+3*0 = $ 54

The optimal solution is 8 chairs and 4 tables.

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Find<br><br><br>(a) the value of q in radians <br><br>(b) the area of the shaded region in cm²

Tresset [83]

Answer:

(a) 1.18

(b) 99.71

Step-by-step explanation:

to know the value of q in degrees we can use cosine of q

$\cos (q) = \frac{OR}{OQ}\\\\\cos (q) = \frac{5}{13}\\\\q = \cos^{-1}(\frac{5}{13})\\\\q \approx 67.38$

now to radians

the formula is

$x\times\frac{2\pi}{360}\\\\$

with x the degrees

$67.38\times \frac{2\pi}{360}\\\\=\frac{67.38\pi}{180}\\\\\approx 0.374\pi\\\\\approx 1.18$

so the measure of angle q is 1.18 radians

so now for part b

$A = \frac{r^2 \alpha }{2}\\\\$

with $\alpha$ being the central angle in radians

for degrees is the following

$A = \frac{\theta}{360}\times \pi r^2$

so we have

$A = \frac{13^2 (1.18)}{2}\\\\A = 99.71$ cm^2

4 0

3 years ago

The number of students taking algebra increased 20% in one year. If

Fiesta28 [93]

Answer: 50 students

Step-by-step explanation:

You know that the number of students last year plus 20% of that number equals the number of students this year, which can be written as: x + 0.2x = 60 →

1.2x = 60 → x = 50 students :)

7 0

2 years ago

Suppose a shoe factory produces both low-grade and high-grade shoes. The factory produces at least twice as many low-grade as hi

umka2103 [35]

Answer:

The factory should produce 166 pairs of high-grade shoes and 364 pairs of low-grade shoes for maximum profit

Step-by-step explanation:

The given parameters for the shoe production are;

The number of low grade shoes the factory produces ≥ 2 × The number of high-grade shoes produced by the factory

The maximum number of shoes the factory can produce = 500 pairs of shoes

The number of high-grade shoes the dealer calls for daily ≥ 100 pairs

The profit made per pair of high-grade shoed = Birr 2.00

The profit made per of low-grade shoes = Birr 1.00

Let 'H', represent the number of high grade shoes the factory produces and let 'L' represent he number of low-grade shoes the factory produces, we have;

L ≥ 2·H...(1)

L + H ≤ 500...(2)

H ≥ 100...(3)

Total profit, P = 2·H + L

From inequalities (1) and (2), we have;

3·H ≤ 500

H ≤ 500/3 ≈ 166

The maximum number of high-grade shoes that can be produced, H ≤ 166

Therefore, for maximum profit, the factory should produce the maximum number of high-grade shoe pairs, H = 166 pairs

The number of pairs of low grade shoes the factory should produce, L = 500 - 166 = 334 pairs

The maximum profit, P = 2 × 166 + 1 × 364 = 696

6 1

3 years ago

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olya-2409 [2.1K]

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5 0

3 years ago

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Nina [5.8K]

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8 0

3 years ago