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4 years ago

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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

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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To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

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$x = -1$ is local maximum, $x = 1$ is local minimum.

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