With calculus like this, who needs inequality formulas?

Calculus Level 3

A company bought a machine with one million dollars. It cost electricity worth five thousand dollars to run the machine. The maintenance fee is twenty thousand for the first year and increases by twenty thousand every following year . The company should purchase new machine how many years after in order to have the lowest annual machine expense?


The answer is 10.

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2 solutions

Richard Desper
Aug 29, 2019

Let's suppose we replace the machine after n n years. Then the average annual cost of the purchase of the machine will be (in dollars) 1 , 000 , 000 n \frac{1,000,000}{n} .

The electrical cost of the machine is $ 5000 \$5000 /year no matter what value n is given.

As for the maintenance fee, it increases linearly, from $ 20 , 000 \$20,000 the first year to $ 20 , 000 n \$20,000*n the final year, so its average annual cost will be (in dollars) 20 , 000 ( n + 1 ) 2 = 10 , 000 n + 5000 \frac{20,000(n+1)}{2} = 10,000n + 5000 .

So we are seeking to minimize A ( n ) = 1 , 000 , 000 n + 5000 + ( 10000 n + 5000 ) A(n) = \frac{1,000,000}{n} + 5000 + (10000n + 5000) .

Consider the derivative, A ( n ) = 1 , 000 , 000 n 2 + 10000. A'(n) = \frac{-1,000,000}{n^2} + 10000.

Observe A ( n ) < 0 A'(n) < 0 for n < 10 n < 10 , A ( 10 ) = 0 A'(10) = 0 , and A ( n ) > 0 A'(n) > 0 for n > 10 n > 10 . Thus the average annual cost is minimized at n = 10 n=10 .

Kevin Xu
Aug 29, 2019

I believe you guys can do this question in calculus for sure. Write the equation and find the derivative. Make the derivative equal to zero and find out critical points bla bla bla ... \\

Ever think about doing this question with inequality equations? \\ y = 1000 + 5 x + x ( 10 x + 10 ) x = 10 x + 1000 x + 15 y = \frac {1000+5x + x(10x+10)}{x} = 10x + \frac {1000}{x} + 15 \\ 10 x + 1000 x + 15 2 10 x 1000 x + 15 = 215 10x + \frac {1000}{x} + 15 \geq 2\sqrt {10x \cdot \frac {1000}{x}} + 15 = 215 thousand dollars per year. [AM-GM] \\ Two sides equal only when 10 x + 1000 x = 200 > x = 10 10x+\frac {1000}{x} = 200 --> \boxed{x = 10}

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