Mathematics

Mathematics

Problem

The power output (in watts) of a certain machine after tt hours of operation is P(t)=200t−t2P(t)=200t-t^{2}, while the energy consumption rate (in watts) is C(t)=100+2tC(t)=100+2t. The operational efficiency is defined as E(t)=P(t)C(t)E(t)=\dfrac{P(t)}{C(t)}. For what value of tt (in hours) is the efficiency maximized during the interval 0<t<1000<t<100? a) 4141 b) 62 (approximately)62\text{ (approximately)} c) 5050 d) 8282
Intuitive explanation

Efficiency is a ratio of output to input, so maximizing it means finding where E(t)=P(t)C(t)E(t) = \frac{P(t)}{C(t)} reaches its peak. To find that peak, take the derivative using the quotient rule and set it equal to zero—this gives you the critical points where the slope of efficiency changes from positive to negative. Then check which critical point in your interval actually gives the maximum value.

Intuitive explanation

Efficiency is a ratio of output to input, so maximizing it means finding where E(t)=P(t)C(t)E(t) = \frac{P(t)}{C(t)} reaches its peak. To find that peak, take the derivative using the quotient rule and set it equal to zero—this gives you the critical points where the slope of efficiency changes from positive to negative. Then check which critical point in your interval actually gives the maximum value.

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