# Expenses of Production

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In the past area, we took a gander at creation. . In this segment, we take a gander at the expense of generation

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﻿Expenses of Production

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In the past area, we took a gander at creation. In this segment, we take a gander at the cost of creation & deciding the ideal level of yield. We'll begin with a fascinating cost illustration, & then concentrate on deciding the ideal level of yield when all is said in done.

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Optimal Level of Law Enforcement & Crime Prevention. Consider the aggregate cost of wrongdoing (C T ), including both the cost of the criminal demonstration itself (C A ) & the cost of law requirement & wrongdoing counteractive action (C P ). We'd jump at the chance to know the ideal level of law requirement & wrongdoing counteractive action (L) that will limit the aggregate cost C T = C P + C A . Keep in mind from analytics that to limit a capacity, we take the primary subordinate and set it equivalent to zero. So we will have dC T/dL = dC P/dL + dC A/dL = 0 or dC P/dL = – dC A/dL . This implies the ideal implementation level is the place the cost of keeping an extra wrongdoing is equivalent to the cost of an extra criminal act. To complete the process of unraveling for the ideal level (L), we'd have to know the particular type of the cost capacities included.

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Isocost Curve The arrangement of mixes of sources of info that cost a similar sum

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Equation of an isocost Suppose you have 2 inputs, capital K & work L. The cost of a unit of capital is P K . The cost of a unit of work is P L . Give a specific cost a chance to sum be R. At that point all blends of K & L to such an extent that P L L+ P K = R lie on the isocost bend related with that expense. In the event that we rework the condition as K = R/P K – (P L/P K )L , we see that the incline of the isocost is – (P L/P K ) & the vertical capture is R/P K .

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Graph of an isocost For instance, assume you're keen on the expense sum \$10,000. Assume likewise that Labor cost \$10 per unit & capital cost \$100 per unit. At that point the slant of the isocost is – P L/P K = – 10/100 = – 0.1 . The vertical capture would be 10,000/100 = 100 & the even catch is 10,000/10 = 1,000. K R/P K = 100 slant = – P L/P K = – 0.1 R/P L = 1000 L

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Maximizing yield for a given cost level At focuses A & B, we're going through the expense related with this isocost, however we're not creating as much as we can. We're just making Q 1 units of yield. We can't deliver Q 3 or Q 4 with this cost. Those yield levels would cost more. At point E, we're creating the most for the cash, where the isocost is digression to an isoquant. isoquants K An isocost E Q 4 Q 3 Q 2 B Q 1 L

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At the intersection of the isocost & isoquant, the slants of those bends are equivalent. We discovered already that the incline of the isoquant is – MP L/MP K , & the slant of the isocost is – P L/P K . So at the intersection, – MP L/MP K = – P L/P K or, duplicating by - 1, MP L/MP K = P L/P K . This expression is equal to MP L/P L = MP K/P K .

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MP L/P L = MP K/P K This condition implies that to get the most yield for your cash, you ought to utilize information sources with the end goal that the negligible item per dollar is equivalent for all data sources. (See the similitude to the utility expansion condition that the peripheral utility per dollar is equivalent for all products.)

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Minimizing cost for a given yield level At focuses A & B, we're delivering the sought amount, yet we're not utilizing the least expensive mix of information sources, so we're spending more than should be expected. We can't create the sought yield level at cost level C 1 . We require more cash. At point E, we're creating the coveted yield at the most reduced cost, where the isoquant is digression to an isocost. K An E isoquant Q 1 B L isocosts C 1 C 2 C 3

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So whether we're expanding yield for a given cost level, or limiting expense for a given yield level, the condition is the same: MP L/P L = MP K/P K The minor item per dollar is equivalent for all data sources.

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Short Run Costs of Production

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Total Fixed Cost (TFC) Total settled cost is the cost related with the settled info.

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\$ TFC Quantity Since TFC is consistent, its chart is a level line.

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Average Fixed Cost (AFC) AFC = TFC/Q AFC is the settled cost per unit of yield.

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The AFC bend slants descending & gets nearer & nearer to the level hub. \$ AFC Quantity

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Total Variable Cost (TVC) Total variable cost is the cost related with the variable info.

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TVC \$ Quantity The TVC bend is upward slanting. It is frequently drawn like a flipped over S, first getting compliment & compliment, & then more extreme & more extreme. This shape mirrors the expanding & then diminishing peripheral returns we talked about in the area on generation.

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Average Variable Cost (AVC) AVC = TVC/Q AVC is the variable cost per unit of yield.

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We can decide the state of the AVC bend in view of the state of the normal item bend (AP). Assume X is the measure of variable info & P X is its cost. At that point, AVC = TVC/Q = (P X)/Q = P (X/Q) = P X [1/(Q/X)] = P X [1/AP]. So since AP had a transformed U-shape, AVC must have a U-shape.

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Average Variable Cost \$ AVC Quantity

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TC \$ TFC Total Cost TC = TFC + TVC The TC bend resembles the TVC bend, however it is moved up, by the measure of TFC. Amount

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Average Total Cost Like AVC, ATC is U-molded, yet it achieves its base after AVC achieves its base. This is on the grounds that ATC = AVC +AFC & AFC keeps on falling & pulls down ATC. ATC \$ AVC Quantity

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Marginal Cost (MC) MC is the extra cost related with an extra unit of yield. MC = Δ TC/Δ Q Alternatively, MC = dTC/dQ . MC is the main subordinate of the TC bend or the incline of the TC bend.

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We can decide the state of the MC bend in light of the state of the minimal item bend (MP). Assume the firm takes the costs of contributions as given. At that point, MC =  TC/ Q = P X  X/ Q = P X [1/(  Q/ X)] = P X [1/MP]. So since MP had a rearranged U-shape, MC must have a U-shape.

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MC \$ Quantity While MC is U-formed, it is regularly drawn so it stretches out up higher on the correct side.

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MC ATC \$ AVC Quantity Important Graphing Note: T he MC must converge the ATC at its base & the AVC bend at its base.

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We have a comparative graphical elucidation of ATC to the one we had for AP. Since ATC = TC/Q, the ATC of a specific estimation of Q 1 can be deciphered as the slant of the line from the root to the comparing point on the bend. TC  TC 1  0  Q 1 → Q

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We likewise have comparable graphical elucidation of MC to the ones we had for MP. The persistent MC is the slant of the aggregate cost bend at a specific point. The discrete MC is the slant of the line portion interfacing 2 focuses on the aggregate cost bend. TC Q

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TC TR TC Q Breaking Even Recall that TR = PQ. On the off chance that the cost of yield is settled for the firm (concerning a consummately focused firm), then TR is a straight line with slant P. At the point when the TR bend is over the TC bend, the firm will have positive financial benefits. At the point when the TC bend is over the TR bend, the firm will have financial misfortunes. The firm will make back the initial investment (have zero financial benefits) where TR=TC.

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TC TR TC Profit-augmenting yield level Q Maximizing Profit The firm will have the most extreme benefits where the vertical separation between TR & TC is the biggest (& TR is above TC). This is likewise where MR = MC (which you ought to review from Micro Principles is the benefit augmenting condition). That implies that the incline of the TR line meets the slant of the TC bend. So the TR line will be parallel to a digression to the TC line at the point where benefits are augmented.

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TC TR TC Profit-min imizing yield level Q Minimum Profit Notice that the TR line is additionally parallel to a digression to the TC line here. TR – TC achieves a base here, not a greatest.

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will diverge a little to survey from Calculus how to utilize first and second subordinates to decide minima and maxima. y Consider a capacity y = f(x) as appeared. See that it has a base an incentive at x 1 . See additionally that the slant of the capacity (which is the same as the slant of the line digression to the bend by then) is zero. That is, f  (x 1 ) = 0. x 1 x

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y f  (x) > 0 f  (x) < 0 Just to one side of x 1 , the bend slants descending; it has a negative incline. To one side of x 1 , the bend slants upward; it has a positive slant. x 1 x

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y f  (x) < 0 f  (x) > 0 So as we move from left to ideal in the region of x 1 , the slant is going from negative to zero to positive. It is expanding. Review that if a capacity is expanding, its subsidiary is certain. For this situation, the capacity itself is the slant or first subsidiary. So its subsidiary is the second subordinate. At that point, on the grounds that the primary subordinate is expanding, the second subsidiary must be sure: f  (x 1 ) > 0. To assemble this: at the very least x 1 , the principal subordinate f  (x 1 ) = 0 and the second subsidiary f  (x 1 ) > 0 . f  (x 1 ) = 0 x 1 x

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Consider rather this capacity. At x 1 , we have a most extreme. The subordinate f  (x 1 ) = 0. Here, as we move from left to ideal in the region of x 1 , the slant is going from positive to zero to negative. The incline is diminishing. In the event that a capacity is diminishing, its subordinate is negative. Again here, the capacity is the slant or