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NeTakaya
4 years ago
9

Carlo and Anita make mailboxes and toys in their wood shop. Each mailbox requires 1 hour of work from Carlo and 4 hours from Ani

ta. Each toy requires 1 hour of work from Carlo and 1 hour from Anita. Carlo cannot work more than 12 hours per week and Anita cannot work more than 24 hours per week. If each mailbox sells for $10 and each toy sells for $5, then what is their maximum possible revenue

Mathematics
1 answer:
Natali [406]4 years ago
5 0

Answer:

$80

Step-by-step explanation:

Let the number of hours required to make a mailbox = x

Let the number of hours required to make a toy = y

Each mailbox requires 1 hour of work from Carlo and 4 hours from Anita.

Each toy requires 1 hour of work from Carlo and 1 hour from Anita.

The table below summarizes the information for ease of understanding.

\left|\begin{array}{c|c|c|c}&$Mailbox(x)&$Toy(y)&$Maximum Number of Hours\\--&--&--&------------\\$Carlo&1&1&12\\$Anita&4&1&24\end{array}\right|

We have the constraints:

x+y \leq 12\\4x+y \leq 24\\x \geq 0\\y \geq 0

Each mailbox sells for $10 and each toy sells for $5.

Therefore, Revenue, R(x,y)=10x+5y

The given problem is to:

Maximize, R(x,y)=10x+5y

Subject to the constraints

x+y \leq 12\\4x+y \leq 24\\x \geq 0\\y \geq 0

The graph is plotted and attached below.

From the graph, the feasible region are:

(0,0), (6,0), (4,8) and (0,12)

At (6,0), 10x+5y=10(6)+5(0)=60

At (4,8), 10(4)+5(8)=80

At (0,12), 10(0)+5(12)=60

The maximum revenue occurs when they use 4 hours on mailboxes and 8 hours on toys.

The maximum possible revenue is $80.

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tresset_1 [31]

Answer:

|-45|

Step-by-step explanation:

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The absolute value function can be definied as:

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Using this definition, we have:

|-21| = -(-21) = 21

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|-45| = -(-45) = 45

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PLEASE SHOW ALL THE STEPS THAT YOU USE TO SOLVE THIS PROBLEM
Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

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Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer:  {x = 1 , y=1, z=0

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