- A FARMER FINDS THAT IF SHE 
PLANTS 50 TREES PER ACRE,
EACH TREE WILL YIELD 65 BUSHELS 
OF FRUIT.
SHE ESTIMATES THAT FOR EACH 
ADDITIONAL TREE
PLANTED PER ACRE, THE YIELD FOR 
EACH TREE WILL DECREASE
BY THREE BUSHELS.
HOW MANY TREES SHOULD 
SHE PLANT PER ACRE
TO MAXIMIZE HER HARVEST?
SO THE TOTAL HARVEST PER ACRE 
IS GOING TO BE EQUAL
TO THE NUMBER OF TREES x THE 
NUMBER OF BUSHELS PER TREE.
SO IF WE LET X EQUAL THE NUMBER 
OF ADDITIONAL TREES
WE COULD SAY H OF X, 
THE HARVEST PER ACRE,
WOULD BE EQUAL TO THE TOTAL 
NUMBER OF TREES,
WHICH WOULD BE 50 + X x THE 
NUMBER OF BUSHELS PER TREE.
WELL, THE BUSHELS PER TREE START 
AT 65,
BUT IT DECREASES BY 3 BUSHELS 
FOR EXTRA TREE.
SO THE TOTAL NUMBER OF BUSHELS 
PER TREE WOULD BE 65 - 3X.
SO 50 + X IS THE TOTAL 
NUMBER OF TREES AND 65 MINUS 3X
IS THE TOTAL NUMBER OF BUSHELS 
PER TREE.
LET'S GO AHEAD 
AND MULTIPLY THIS OUT.
H OF X = 50 x 65 = 3,250 - 50 x 
3X = - 150X + 65X - 3X SQUARED.
LET'S GO AHEAD AND COMBINE 
LIKE TERMS
AND WRITE THE TERMS 
IN DESCENDING ORDER.
WE'D HAVE H OF X = -3X SQUARED,
COMBINE THE X TERMS WOULD GIVE 
US - 85X + 3,250.
SO OUR GOAL HERE IS TO MAXIMIZE 
THIS FUNCTION.
SO FOR THE NEXT STEP WE'LL 
DETERMINE THE CRITICAL NUMBERS,
WHICH IS WHERE THE FIRST 
DERIVATIVE
IS EQUAL TO ZERO OR UNDEFINED.
SO H PRIME OF X = -6X - 85 = 0.
NOTICE HOW THE DERIVATIVE 
FUNCTION IS A LINEAR FUNCTION,
WHICH IS NEVER UNDEFINED,
SO TO FIND THE CRITICAL NUMBER 
WE'LL SET THIS EQUAL TO ZERO
AND SOLVE FOR X.
SO WE'LL ADD 85 TO BOTH SIDES,
-6X EQUAL 85 
DIVIDE BOTH SIDES BY -6
AND WE HAVE X 
IS APPROXIMATELY -14.17.
AND SINCE X IS THE NUMBER 
OF ADDITIONAL TREES,
WE'RE GOING TO ROUND TO THE 
NEAREST WHOLE NUMBER.
SO WE'LL USE THE CRITICAL NUMBER 
X EQUALS -14,
BUT WE SHOULD STILL VERIFY THAT 
THE FUNCTION IS MAXIMIZED
NOT MINIMIZED AT THIS VALUE.
WHAT WE SHOULD RECOGNIZE THAT 
THE HARVEST FUNCTION HERE
IS A QUADRATIC FUNCTION WITH A 
NEGATIVE LEADING COEFFICIENT.
SO WE HAVE A PARABOLA 
THAT OPENS DOWN,
OR A CONCAVE DOWN FUNCTION,
THEREFORE THIS X VALUE 
WOULD MAXIMIZE THE FUNCTION
OR WE CAN ALSO USE THE FIRST 
OR SECOND DERIVATIVE TEST.
LET'S USE THE SECOND DERIVATIVE 
TEST TO VERIFY THE CONCAVITY.
SO HERE'S THE FIRST DERIVATIVE 
FUNCTION
AND NOTICE HOW THE SECOND 
DERIVATIVE FUNCTION,
H DOUBLE PRIME OF X, 
WOULD BE EQUAL TO -6.
BECAUSE THE SECOND DERIVATIVE 
IS ALWAYS NEGATIVE,
THE FUNCTION IS ALWAYS CONCAVE 
DOWN,
WHICH IS GOOD NEWS 
BECAUSE THAT MEANS
THAT OUR CRITICAL NUMBER WE HAVE 
A MAXIMUM FUNCTION VALUE
AND THEREFORE THE FUNCTION 
IS MAXIMIZED
AT APPROXIMATELY X EQUAL -14.
BUT WE MIGHT BE THINKING 
TO OURSELVES,
HOW CAN X BE NEGATIVE IF X IS 
THE NUMBER OF ADDITIONAL TREES?
WELL, THE TOTAL NUMBER OF TREES 
IS 50 + X,
SO THE IDEAL NUMBER OF TREES 
WOULD BE 50 + -14
OR 36 TREES PER ACRE,
SO 50 TREES IS TOO MANY TO 
MAXIMIZE THE HARVEST.
SO THE HARVEST IS MAXIMIZED WITH 
50 + X TREES,
WHICH IS 50 + -14 OR 36 TREES.
I HOPE YOU FOUND THIS HELPFUL.
