A shoe manufacturer determines that its monthly revenue, R(q)=−0.31(q−260)^2+9500 in dollars, is given by the function defined above, where qqq is the number of pairs of shoes sold each month. What is the maximum value of the company's monthly revenue in dollars?

It should be noted that maximum value of the company's monthly revenue is $9500.

Therefore, the maximum value of the company's monthly revenue in dollars will be $9500.

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The maximum value of the company's monthly revenue is $9500.

Given to us,

shoe manufacturer monthly revenue =

where, q is the number of pairs of shoes sold each month,

To maximize the value of the company's monthly revenue, we need to differentitae the function,,

After diffrentiating it for the first time we get,

equating it against 0,

we get, q = 260,

Equating the value of q into the function we will get $9500.

Therefore, the maximum value of the company's monthly revenue is $9500, this happened because -0.31(q-260)² is negative we need to remove that value to get the maximum value.

Hence, the maximum value of the company's monthly revenue is $9500.

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answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

hope this ^_^

step-by-step explanation: