3.When production is 2100, marginal revenue is 6.75 dollars per unit and margina

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3.When production is 2100, marginal revenue is 6.75 dollars per unit and marginal cost is 7.75 dollars  per unit. Do you expect maximum profit to occur at a production level above or below 2100? (Enter above or below.)

height =

Consider the function f(x)=3x3-4x on the interval [-4,4] Find the average or mean slope of the function on this interval. By the Mean Value Theorem, we know there exists at least one c in the open interval (-4,4) such that f'(c) is equal to this mean slope. For this problem, there are two values of c that work. The smaller one is and the larger one is Consider the function f(x)=4-8x2 on the interval [-3,8] Find the average or mean slope of the function on this interval, i.e. f(8)-f(-3)/8-(-3)= By the Mean Value Theorem, we know there exists a c in the open interval (-3,8) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it. .When production is 2100, marginal revenue is 6.75 dollars per unit and marginal cost is 7.75 dollars per unit. Do you expect maximum profit to occur at a production level above or below 2100? If production is increased by 80 units, what would you estimate the change in profit would be? dpi A ball that is dropped from a window hits the ground in 7 seconds. How high is the window? (Give your answer in feet; note that the acceleration due to gravity is 32 ft/s 2 ) height = Suppose that the demand of a certain item is x=10+1/p2 Evaluate the elasticity at 0.7: E(0.7)

Explanation / Answer

can find the average slope on the interval by looking at change in y / change in x at the end points.

avg slope = f(5)-f(3) / 5 - 3 = 2sqrt(5)+3-2sqrt(3)-3 / 5-3 = 2 ( sqrt(5)-sqrt(3)) / 2 = sqrt(5) - sqrt(3)

Then you are looking for the point c in this interval where the actual slope is equal to this value. In other words you want the point where the tangent line to the curve is parallel to the line between the two points (2, f(2)) and (5 , f(5)).

Take the derivative of f(x) and set it equal to the value we got for avg slope,

f(x) = 2x^1/2 +3 ( I like to rewrite the sqrt first. It helps when differentiating)
f'(x) = 2(1/2)x^-1/2 = x^-1/2

so want sqrt(5) - sqrt(3) = x^-1/2
sqrt(5) - sqrt(3) = 1/sqrt(x)
sqrt(x) = 1/(sqrt(5)-sqrt(3))
x = (1 / (sqrt(5) - sqrt(3)))^2
x = 3.9365...
Obviously since we are dealing with square roots of primes this is an approximate value. If you need the exact just leave it in the second to last form. That's as simplified as you're going to get.

An easy check is to make sure your value is actually in the interval. You can also double check by plugging this back into the derivative formula you solved for and check it against the avg slope. They should be equal.

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