Question

26. In the process of producing 1 hormone-free, stress-free, happy cow, the New Age Cattle Ranch produces I ton of natural, organic plant fertilizer. Demand for the ranch's cattle is given by Pe -2,000 5Q, and demand for fertilizer is given by P 500 4Q. The cost of producing Q cow-fertilizer bundles is C- 5,000+4000. To maximize profits, the ranch should a. produce 160 cow-fertilizer units, sell 160 cattle, and sell 62.50 tons of fertilizer. b. produce 160 cow-fertilizer units, sell 160 cattle, and sell 12.50 tons of fertilizer. c. produce 160 cow-fertilizer units, sell 160 cattle, and sell 160 tons of fertilizer. d. produce 116.67 cow-fertilizer units, sell 116.67 cattle, and sell 116.67 tons of fertilizer. e. produce 200 cow-fertilizer units, sell 200 cattle, and dispose of all the fertilizer in an environmentally friendly manner.

Answer 1

Given, $\large P_c = 2000 - 5Q_c$

Thus, total revenue will be

$\large TR_c=P_c Q_c= (2000 - 5Q_c)Q_c$

$\large TR_c = 2000Q_c - 5Q_c^2$

Now differentiate wrt Qc we get

$\large MR_c = 2000 - 10Q_c$

Similarly, the fertilizer Marginal revenue can be determined

$\large P_f = 500 - 4Q_f$

$\large TR_f = 500Q_f - 4Q_f^2$

Differentiating above equation wrt Qf we get

$\large MR_f = 500 - 8Q_f$

The total cost function of the firm is given by

C = 5000 + 400Q

MC = 400

In case of cow equate MR WITH MC

$\large 2000 - 10Q_c = 400$

$\large 10Q_c = 2000 - 400$

$\large Q_c = 160$

$\large P_c = 2000 - 5*160 = 2000-800 = \$ 1,200$

Similarly equate fertilizers MR to MC we get

$\large 500 - 8Q_f = 400$

$\large 8Q_f =500 - 400$

$\large Q_f =\frac{100}{8} = 12.5 units$

$\large P_f = 500 - 4Q_f = 500 - 4*12.5 = \$ 450 per unit$

Option B is correct.

Produce 160 cow-fertilizer units, sell 160 and sell 12.5 tons of fertilizer.

Please contact if having any query thank you.