# math

A trucker hauls citrus fruit from Florida to Montreal. Each crate of oranges is 4 ft3 in volume and weighs 60 lb. Each crate of grapefruit has a volume of 6 ft3 and weighs 75 lb. His truck has a maximum capacity of 300 ft3 and can carry no more than 4200 lb. Moreover, he is not permitted to carry more crates of grapefruit than crates of oranges. If his profit is \$2.50 on each crate of oranges and \$4 on each crate of grapefruit, how many crates of each fruit should he carry for maximum profit?

oranges

grapefruits

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1. orange crates = x
grapefruit crates = y

constraints
a)4x+6y</= 300
b)60 x + 75 y</= 4200
c)y</= x

maximize p = 2.5 x + 4 y

graph those constraint lines and mark intersections

a hits c at (30,30)
b hits c at (31.11, 31.11)
a hits b at (45,20)

our region is between the x axis and those lines, check intersections for max p
p = 2.5 x + 4 y
(30,30) --> p = 195
(31.11,31.11) --> p = 202.2
(45,20) -> p = 192.5
(70,0) -> P = 175

SO 45 ORANGE, 20 GRAPEFRUIT

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2. oh, if you graphed carefully, no need to do (31.11,31.11) because it is above others. however the (70,0) point where b hits the x axis should be checked even if you can see it is not going to be max.

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