Good Morning! 1) Office hrs: 1:30-2:30 in MSC 2:30-3 in PDL C-326 2) Homework 3 is due tomorrow (WS 6, 7, 8) 3) Midterm 1 is next Tuesday (Prologue + WS 1-9) Reviews in class: Thursday & Friday (Please print last quarter’s midterms from our class website and bring to quiz section tomorrow) Out-of-class reviews: listed on website.

Recap of last time (Worksheet 8) Given a specific Total Cost vs. quantity produced q, recall: 1.The Breakeven Price (BEP): Is the special price that satisfies: If p < BEP, the producer never makes a profit If p > BEP, the producer makes some profit, for some quantities

0 TC Breakeven Price = slope of the lowest diagonal line tangent to TC Draw this line & Pick a good point on this line: (450, 1100) Our BEP =1100/450=2.44 ($ per hat) hats $ Recall Computing BEP:

hats $ Slope of this line= TC(50)/50= =AC(50) Slope of this line= TC(200)/200= AC(200) (lower than previous!) Slope of this line= TC(400)/400= AC(400) (lower still!) Note: BEP = minimum AC

Also, recall: 2.The Shutdown Price (SDP): Is the special price that satisfies: If p < SDP, the producer never makes a profit AND loses more money than the fixed costs if he/she produces any hats. One should shut down and produce nothing! If p > SDP, the producer will at least recover some of the fixed cost FC, for some quantities, so it’s best to stay open a while.

0 VC Shutdown Price = slope of the lowest diagonal line tangent to VC Draw this line & pick a good point on this line: (600, 900) Our SDP =900/600=1.5 ($ per hat) hats $ Recall: Computing SDP:

0 VC Similarly to the situation for BEP (where BEP= min AC) we also have: hats $ Shutdown Price = slope of the lowest diagonal line tangent to VC, so SDP= minimum AVC

0 TC hats $ TR line for a price p between SDP and BEP If zero hats are made, the loss = -FC = - $600 If 750 hats are made, the loss = -$200

0 TC hats $ TR line for a price p below SDP If zero hats are made, the loss = -FC = - $600 If hats are produced, the loss is always larger than $600!

Worksheet 9: Analysis of Cost (Part II)

First, draw TR for a market price of p=$2.50 per paperweight. slope=2.5=2.5/1=25/10=250/100=500/200  a good point on the line is (200, 500) TR

Method 1: Given the graphs of TR and TC recall that the max profit occurs  where we see the greatest max distance between the graphs of TR and TC, (with TR on top). Rolling ruler: hold ruler vertical and move across, searching for the largest “gap”. Part I of WS 9: Three methods to compute max profit, from graphs: Looks like max profit occurs at about q=650 paperweights.

Preparation for Methods 2 & 3: From WS 3, recall that our profit is maximized at the first quantity where we go from MR>MC (increasing profit) to MR<MC (decreasing profit). So we’re looking for q where MR=MC. In our case, MR=$2.50, always, because you make $2.50 from selling any extra item. Note: If we are in a market price situation (each item sells for $p per item), then TR is a diagonal line of slope p and MR(q)=p for every value of q. How about MC? Recall that MC=ΔTC if Δq=1, and we can think of MC as the slope of a secant line through the TC graph at points q and q+1. (because slope=rise/run= ΔTC / Δq= ΔTC/1=MC) Problem: The scale of our graph is too big to “see” a Δq of 1 paperweight! Solution: If the scale for the x-axis is in tens, hundreds, or thousands of items, we can use the slope of the tangent line at q (instead of the secant) to get MC(q)

? 2. Pick 2 easy to read points:(350, 1000) & (750, 1250) MC (200)=? 1. Draw the tangent line at q= Example:

Method 2 (given the graphs of TR and TC): The max profit occurs when MR=MC (switching from MR>MC to MR<MC) i.e. when the graphs for TR and TC have parallel tangent lines (since “matching slopes”  MR=MC)

Method 2: look for matching slopes (parallel tangent lines) at the same q. In our example, since TR is already a straight line, its slope (MR) is always 2.5. Align the ruler with TR and move it parallel until it becomes tangent to TC Find the quantity q where the slopes match (MR=MC): Once again, looks like max profit occurs at about 650 paperweights.

Method 3: If you’re given the graphs of MR and MC, simply look for their intersection point. Note: If MR greater than MC before, and smaller after, that q gives max profit. Otherwise it gives max loss. MR So max profit is at q=640 (note: more accurate answer!) MR=MC MR(q)=2.5 Look at your handout: The graph of MC is given. How do we draw the graph of MR? MR is always 2.5