The square on the right is a scaled copy of the square on the left identify the scale factor

A farmer owns a 100 acre farm and plans to plant at most three crops. The seed for crops A,B, and C costs $40, $20, and $30 per

aev [14]

Answer:

The number of acre per crop for maximum profit is;

Crop A = 0 acres

Crop B = 60 acres

Crop C = 40 acres

Profit, P = $26,000

Step-by-step explanation:

We plant A in X acres

B in Y acres and

C in Z acres

Therefore, X + Y + Z ≤ 100

we work A for 1·X workdays

B for 2·Y workdays and

C for 1·Z workdays

Where X + 2·Y + Z ≤ 160

$40·X for crop A

$20·Y for crop B and

$30·Z for crop C is spent whereby

$40·X + $20·Y + $30·Z ≤ $3200

The farmer makes

$100·X from crop A

$300·Y from crop B and

$200·Z from crop C

P = $100·X + $300·Y + $200·Z

Therefore, we have three equations with three unknowns solving the equations simultaneously, we have

X + Y + Z = 100....................................................(1)

X + 2·Y + Z = 160.................................................(2)

$40·X + $20·Y + $30·Z = $3200....................(3)

By subtracting equation (1) from (2) gives Y = 60 acres

Multiplying equation (1) by 40 and subtracting from (3) we have Z = -40

and therefore, Y = 80

If he plants only B the farmer has 2 work day per acre and since there is a max of 160 days, he can only plant on 80 acres, therefore, total profit = $300 × 80 = $24,000

Comparing the profit per acre to the seed cost we have Profit for seed A = $100 while cost = $40 per acre,

If we remove seed A we have

Y + Z = 100....................................................(4)

2·Y + Z = 160.................................................(5)

$40·X + $20·Y + $30·Z = $3200....................(3)

Solving equation (4) and (5), we have Y = 60 acres and Z = 40 acres

Therefore, the profit becomes

$300 × 60 + $200 × 40 = $26,000.

3 0

2 years ago

Factor the cubic polynomial 6x3 – 11x2 – 12x 5. use the rational root theorem and synthetic division

pochemuha

The factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is:

6x³ - 11x² - 12x + 5 = (x + 1)(x - $\frac{5}{2}$ )(x - $\frac{1}{3}$ )

For given question,

We have been given the cubic polynomial 6x³ - 11x² - 12x + 5

We need to factorize given cubic polynomial.

By the rational roots theorem, any rational zero of f(x) is expressible in the form ± $\frac{p}{q}$ for integers p, q with p a divisor of the constant term 5 and q a divisor of the coefficient 6 of the leading term.

Factors of p = 5: 1, 5

Factors of q = 6: 1, 2, 3, 6

That means that the only possible rational zeros are:

±{ $\frac{1}{1} ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{6} ,\frac{5}{1} ,\frac{5}{2} ,\frac{5}{3} ,\frac{5}{6}$ }

= ±{ $1 ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{6} ,5 ,\frac{5}{2} ,\frac{5}{3} ,\frac{5}{6}$ }

We need to find the exact zeros of given cubic polynomial.

For x = 1,

6(1)³ - 11(1)² - 12(1) + 5 = -12

This means, x = 1 is not a zero of given cubic polynomial.

For x = -1,

6(-1)³ - 11(-1)² - 12(-1) + 5 = 0

This means, x = -1 is a zero of given cubic polynomial and (x + 1) is a factor.

To factorize given cubic polynomial we use synthetic division.

The synthetic division (6x³ - 11x² - 12x + 5) ÷ (x + 1) is as shown in following image.

⇒ 6x³ - 11x² - 12x + 5 = (x + 1)(6x² - 17x + 5)

The factors of above quadratic polynomial 6x² - 17x + 5 are:

⇒ 6x² - 17x + 5 = (x - $\frac{5}{2}$ )(x - $\frac{1}{3}$ )

So, the factors of given cubic polynomial are:

⇒ 6x³ - 11x² - 12x + 5 = (x + 1)(x - $\frac{5}{2}$ )(x - $\frac{1}{3}$ )

Therefore, the factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is: 6x³ - 11x² - 12x + 5 = (x + 1)(x - $\frac{5}{2}$ )(x - $\frac{1}{3}$ )

Learn more about the polynomial here:

brainly.com/question/16594008

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