Question 1

Differentiate the following function with respect to Q: TC = 50 + 100Q -6Q² +.5Q³ A)dTC/dQ = 50 + 100 -6Q + .5Q² B)dTC/dQ = 100 -12Q + 1.5Q² C)dTC/dQ = 50 + 100 -2Q + 3Q² D)dTC/dQ = 100

Accepted Answer

When differentiating TC with respect to Q, all constant terms (such as 50) are dropped since they do not vary with Q. The derivative of 100Q is 100, and the derivative of the third term, -6Q², is -6² since Q is being differentiated and ² is a constant. Similarly, the derivative of the last term, .5Q³, is .5³. Therefore, the final derivative is: dTC/dQ = 100 -6² + .5³. Simplifying, we get: dTC/dQ = 100 -12Q + 1.5Q², which matches choice B.

Question 2

The total revenue function (where Q = output),is: TR = 400 Q - 4 Q² A)TR is maximized at Q = 20 B)TR is maximized at Q = 30 C)TR is maximized at Q = 40 D)TR is maximized at Q = 50 E)TR is maximized at Q = 60

Accepted Answer

D

Question 3

If the first derivative of Y with respect to X is: dY/dX = -4·X²,then the second derivative is: A)-4 B)-8•X C)-4•X D)-8·X² E)-8

Accepted Answer

The second derivative is the derivative of the first derivative. Differentiating -4·X^2 with respect to X gives -8·X.

Question 4

The following is a cubic demand function in P.Find the derivative $\textbf{ dQ/dP }$ of: Q= 4+3P-.5P² + .02P³. A)dQ/dP = 4 + 3P - P + .06P² B)dQ/dP = 3 C)dQ/dP = 3 - P + .06P² D)dQ/dP = .06P² E)dQ/dP = .06

Accepted Answer

Taking the derivative of the demand function Q with respect to P, we get: dQ/dP = 3 - P + .06P² Therefore, the correct answer is C.

Question 5

The second derivative of the function (d²Y/dX² )is negative at the optimal solution of X=22.Therefore,we know that the solution X=22,where the first derivative equals zero… A)must be a minimum. B)must be a maximum. C)may be either a maximum or a minimum. D)would be nothing,because the second derivative is negative.

Accepted Answer

A negative second derivative indicates that at the critical point (where the first derivative equals zero), the function is concave down. This means that the critical point is a local maximum if the first derivative changes from positive to negative at that point, or a local minimum if the first derivative changes from negative to positive. Since we know that the first derivative equals zero at X=22, and the second derivative is negative, we can conclude that X=22 is a local minimum. Therefore, the best choice is (A) that X=22 must be a minimum.

Question 6

Differentiate the following TC function: TC = 150 + 200 Q - 4 Q² + .6 Q³ A)dTC/dQ = 200 - 8Q + 1.8 Q² B)dTC/dQ =-8 + 1.8 Q² C)dTC/dQ = 200 D)dTC/dQ = 200 - 4Q + .6Q² E)dTC/dQ = 1.8 Q²

Accepted Answer

The derivative of TC with respect to Q is:dTC/dQ = 200 - 8Q + 1.8 Q²

Deck 20: Differential Calculus Techniques in Management