Answer:
The farmer should split 800 for market 1 and 200 for the market 2.
Step-by-step explanation:
The farmer has a total of 1000 apples. He will divide in x apples for the market 1 and y=1000-x for the market 2.
Then, the revenue for the apple sales is the sum of the revenues from market 1 and market 2:
[tex]R=xP_1+yP_2\\\\R=x*2\sqrt{x}+(1000-x)4*\sqrt{1000-x}\\\\R=2x\sqrt{x}+4(1000-x)\sqrt{1000-x}\\\\R=2x^{3/2}+4(1000-x)^{3/2}[/tex]
To maximize the revenue, we derive and equal to zero.
[tex]\dfrac{dR}{dx}=\dfrac{d}{dx}[2x^{3/2}+4(1000-x)^{3/2}]\\\\\\\dfrac{dR}{dx}=2*(3/2)x^{1/2}+4*(3/2)(-1)(1000-x)^{-1/2}=0\\\\\\3x^{1/2}-6(1000-x)^{1/2}=0\\\\\\x^{1/2}=2(1000-x)^{1/2}\\\\x=2^2(1000-x)=4(1000-x)\\\\x=4000-4x\\\\5x=4000\\\\x=4000/5\\\\x=800[/tex]
Then, the quantity y is:
[tex]y=1000-x=1000-800=200[/tex]
