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3 years ago

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Cooperton Mining just announced it will cut its dividend from $4.01 to $2.57 per share and use the extra funds to expand. Prior

to the announcement, Cooperton's dividends were expected to grow at a 3.4% rate, and its share price was $50.07. With the planned expansion, Cooperton's dividends are expected to grow at a 4.7% rate. What share price would you expect after the announcement? (Assume that the new expansion does not change Cooperton's risk.) Is the expansion a good investment?

1 answer:

mina [271]3 years ago

8 0

Answer:

$34.35

The price has fallen from $50.07 to $34.35 which means that Expansion will not be a good option.

Explanation:

Computation for the share price to expect after the announcement

Using this formula

Ke = [ D1 / P0 ] +g

Where,

D1 =$4.01

P0 = $50.07

g =3.4%

Let plug in the formula

Ke = [ D1 / P0 ] +g

Ke= [ $4.01 / $50.07] + 0.034

Ke= 0.0800+ 0.034

Ke= 0.1140

Second step is to find the Price after Expansion using this formula

P0 = D1 / [ Ke - g ]

Where,

D1=$2.57

Ke=0.1140

g=4.7%

Let plug in the formula

P0= $ 2.57 / [ 0.1140 - 0.047 ]

P0=$2.57/0.067

P0=$ 34.35

Based on this calculation, we can see that the price has fallen from $50.07 to $34.35 which means that Expansion will not be a good option.

Therefore the share price that you would expect after the announcement will be $34.35

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Data Given:

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So, we know that Q = $q_{1}$ + $q_{2}$

where $q_{}$ being the individual firm output.

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Including the marginal cost of firm 1 and multiplying the whole equation by $q_{1}$

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Taking the derivative w.r.t to $q_{1}$ , we will get:

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Dividing the whole equation by 4

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Solving for $q_{1}$

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Let's suppose new equation is Y

Y = 1660 $q_{2}$ - 4 $q_{1}$ $q_{2}$ -4 $q_{2} ^{2}$ - 60 $q_{2}$

Pugging in the value of $q_{1}$

Y = 1660 $q_{2}$ - 4 $q_{2}$ (200 - 0.5 $q_{2}$ ) -4 $q_{2} ^{2}$ - 60 $q_{2}$

Y = 1660 $q_{2}$ - 800 $q_{2}$ +2 $q_{2} ^{2}$ -4 $q_{2} ^{2}$ - 60 $q_{2}$

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Taking the derivative w.r.t $q_{2}$

$Y^{'}$ = 1600 - 800 - 4 $q_{2}$ = 0

Solving for $q_{2}$

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Plugging in the value of $q_{2}$ to get the value of $q_{1}$

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