Minimum of a two value function f(r,t)?

A man goes to a large, round market to buy fish. He does not have much money so he wants to

spend as little as possible. His problem is that the fish stands at the edge of the market have

higher prices than those at the centre: the price decreases linearly from the edge to the centre.

To complicate his life, the price at the centre of the market rises with time, and so does the

price in the rest of the market, always changing linearly from edge to centre.

Given that:

• the market is a circle with radius R = 5 km;

• the man can walk at a maximum speed of 2.5 km/h;

• the price at the edge of the market is fixed to Pe = 300 $/kg;

• at the start, t = 0, the price of fish at the centre of the market is Pc0 = 10 $/kg;

• in one hour the price at the centre has risen such that the price is the same in the whole

market: Pf = 300 $/kg.

At what time and distance from the edge of the market will the man be able to buy the cheapest

fish? How many kilograms of fish will he be able to buy if he has 100 $ with him?

2 Answers

  • hfshaw
    Lv 7
    3 years ago
    Favorite Answer

    The cost function (price per kg of the fish) in the market can be written as:

    P(r,t) = Pe - (1 - r/R)*[Pe - Pc(t)]

    where Pc(t) is the time-dependent price at the center of the market. Pc(t) = Pc0 + k*t

    where k = $290/(kg*hr) and 0 < t < 1 hr, so:

    P(r,t) = Pe - (1 - r/R)*[Pe - (Pc0 + k*t)], for 0 < t < 1 hr

    Assuming the man starts at the very edge of the market at time t = 0, and he walks in a straight line directly toward the center of the market at a speed of s = 2.5 km/hr, his position as a function of time is given by: r(t) = R - s*t. We can use this expression to eliminate either r or t in the cost function. Let's eliminate r:

    P(t) = Pe - (1 - (R - s*t)/R)*[Pe - (Pc0 + k*t)]

    P(t) = Pe - (s*t/R)*[Pe - Pc0 - k*t]

    P(t) = Pe - (Pe - Pc0)*s*t/R + k*s*t²/R

    This is the cost of the fish at the location of the man at time, t. This is a quadratic function of time. We want to find the value of t that minimizes this function. As usual, differentiate with respect to t, set the derivative equal to zero and solve for t:

    dP(t)/dt = -(Pe - Pc0)*s/R+ 2k*s*t/R

    0 = -(Pe - Pc0)*s/R + 2k*s*t/R

    (Pe - Pc0) = 2k*t

    t = (Pe - Pc0)/(2k)

    Plugging in the values for the parameters in this case:

    t = ($300/kg - $10/kg)/(2*$290/(kg*hr)) = 0.5 hr, at which time the man has walked 1.25 km (i.e., he is 1.25 km from the market's edge) and the price at that position and time is:

    P(0.5 hr) = $300/kg - (1 - (5km - 2.5km/hr*(hr/2))/(5km))*[$300/kg - ($10/kg + ($290/(kg*hr))*hr/2)] = $263.80/kg

    If he has $100, he can buy $100/($263.80/kg) = 0.379 kg of fish.

  • 3 years ago

    I framed it as a cost function of time t, differentiated to find the minimum (30 mins) and solved for C=290/(60*120)t**2-290/120t. for C(30)=263.75

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