Optimizing Table Delivery: Maximizing Profit and Minimizing Cost

[gillgill]

Your firm offers to deliver 300 tables to a dealer, at $90 per table, and to reduce the price per table on the entire order by 25cent for each additional table over 300.
Find the dollar total involved in the largest possible transaction between the manufacturer and the dealer; then find the smallest possible dollar amount.

i don't know how to start this question.
equations being:
300+x
90-0.25x
?
please help

 

[benorin]

You want to express the cost of the transaction as a function of the number of tables purchased by, say C(t), where t denotes the # of tables purchased and C(t) the cost thereof. We need to determine C(t) explicitly that we may deduce its extrema.

"Your firm offers to deliver 300 tables to a dealer, at $90 per table, and to reduce the price per table on the entire order by 25cent for each additional table over 300."

The easiest way to model the cost function is piecewise:

If 300 or less tables are sold, then the price of each table is $90, so

C(t)=90t,\mbox{ for } t\leq 300

after that, they get cheaper, so we modify the cost function to reflect that change in price per table. For each table sold over 300, the price per table drops $0.25 from the $90 base price per table, that is [$90 - (discount=$0.25 per table)(# of times to apply discount=t-300)] is the price per table if they bought t tables and t>300, then multiply by the # of tables purchased to get the total cost. Hence

C(t)=\left[ 90-0.25\left( t-300\right) \right] t,\mbox{ for } t>300

this can be simplified to C(t)=165t-0.25t^{2},\mbox{ for } t>300

So putting it all togeather, we have the piecewise cost function C(t) given by

C(t)=\left\{\begin{array}{cc}90t,&amp;\mbox{ if }<br /> t\leq 300\\165t-0.25t^{2}, &amp; \mbox{ if } t&gt;300\end{array}\right.

From here, do what you would expect to: find the relative extrema of C(t).

Here is a plot of C(t) I did with maple...

 

[vaishakh]

.
The minmum is very easy. It happens at 660 tables when we get all of them free of cost. The maximum occurs at 330. I think the equation must be 27000 + x(90 - (1/4)x) - 75x. Here x is the extra tables bought after 300. 75x is being subtracted from amount because an amount of 75$ is reduced from the total for the purchase of each table. The derivative is 15 - 1/2x. At this juncture we get the maxima to be

 

[gillgill]

i still don't really get how u get the min. at 660 tables...

 

[benorin]

It's simple.

If, for every table over 300 the price per table decrases by $0.25, then after \frac{90}{0.25}=360 such discounts the tables are free, this occurs for a purchase of 300 + 360 = 660 tables are purchased, 300 to qualify for the discount, 360 more to reduce the price per table to nothing.

The rest of the plot is revealing.

 

[gillgill]

o..ic..but this is a extrema (max/min) question...is there a way to show it using derivatives?

 

[benorin]

Yes, that is how it was done above. For

C(t)=\left\{\begin{array}{cc}90t,&amp;\mbox{ if }t\leq 300\\165t-0.25t^{2}, &amp; \mbox{ if } t&gt;300\end{array}\right.

find the value(s) of t for which \frac{d}{dt} C(t)=0.

Note that cost should not be negative, so look for the extrema of C(t) for t in some bounded interval, namely [0,660] (since values of t outside this range don't make good sense in terms of $ and tables.)

 

[gillgill]

thanks so much.