Math B/6.5
When 14 apple trees are planted per acre in a certain orchard, the average yield per tree is 480 apples per year. For each additional tree planted per acre in the same orchard, the annual yield per tree decreases by 10 apples. How many additional trees should be planted in the orchard so that the maximum number of apples will be produced per year?
When 14 apple trees are planted per acre in a certain orchard, the average yield per tree is 480 apples per year. For each additional tree planted per acre in the same orchard, the annual yield per tree decreases by 10 apples. How many additional trees should be planted in the orchard so that the maximum number of apples will be produced per year?
-
let x extra trees above 14/acre be planted
# of apples N = (14+x)(480 - 10x)
N = -10x^2 +340x + 67200
N' = -20x +340 , and N" = -20, so setting N' to 0 will give a maxima
-20x + 340 = 0
x = 17 extra trees per acre
========== ==========
# of apples N = (14+x)(480 - 10x)
N = -10x^2 +340x + 67200
N' = -20x +340 , and N" = -20, so setting N' to 0 will give a maxima
-20x + 340 = 0
x = 17 extra trees per acre
========== ==========
-
Others have shown the standard derivative solution.
Here's something different:
Apples = (14 + x) (480 - 10x) = (14 + x) (48 - x) * 10
To maximize Apples, make (14 + x) = (48 - x)
so both contribute equally to the product.
14 + x = 48 - x
2x = 34
x = 17
Total trees = 17+14 = 31
Checking:
30 trees: 30 * (480 - 160) = 30 * 320 = 9600
31 trees: 31 * (480 - 170) = 31 * 310 = 9610 ← Maximum apples
32 trees: 32 * (480 - 180) = 32 * 300 = 9600
The reason to make them equal is shown by the following:
Let a * b be a product we want to maximize.
If a != b, then Let c = a + x (where x is some number, positive, negative or zero))
and d = a - x (compensating change for an "area of rectangle" problem
where a and b are to be the width and length and a+b = half the perimeter).
Then cd = a^2 - x^2
and since x^2 >= 0 for all x, cd < ab unless a = b
(i.e. x = 0).
Here's something different:
Apples = (14 + x) (480 - 10x) = (14 + x) (48 - x) * 10
To maximize Apples, make (14 + x) = (48 - x)
so both contribute equally to the product.
14 + x = 48 - x
2x = 34
x = 17
Total trees = 17+14 = 31
Checking:
30 trees: 30 * (480 - 160) = 30 * 320 = 9600
31 trees: 31 * (480 - 170) = 31 * 310 = 9610 ← Maximum apples
32 trees: 32 * (480 - 180) = 32 * 300 = 9600
The reason to make them equal is shown by the following:
Let a * b be a product we want to maximize.
If a != b, then Let c = a + x (where x is some number, positive, negative or zero))
and d = a - x (compensating change for an "area of rectangle" problem
where a and b are to be the width and length and a+b = half the perimeter).
Then cd = a^2 - x^2
and since x^2 >= 0 for all x, cd < ab unless a = b
(i.e. x = 0).
-
(14 + x)(480 - 10x) = 0
6,720 - 140x + 480x - 10x² = 0
10x² - 340x = 6,720
x² - 34x = 672
x² - 17x = 672 + (- 17)²
x² - 17x = 672 + 289
(x - 17)² = 961
x - 17 = 31
x = 48
= 48/2 or 24
Answer: 24 additional apple trees (total per acre, 38 trees).
Check:
= (14 + 24)(480 - 10[24])
= 38(480 - 240)
= 38(240)
= 9,120
6,720 - 140x + 480x - 10x² = 0
10x² - 340x = 6,720
x² - 34x = 672
x² - 17x = 672 + (- 17)²
x² - 17x = 672 + 289
(x - 17)² = 961
x - 17 = 31
x = 48
= 48/2 or 24
Answer: 24 additional apple trees (total per acre, 38 trees).
Check:
= (14 + 24)(480 - 10[24])
= 38(480 - 240)
= 38(240)
= 9,120
-
t= number of additional trees planted
f(t)=(14+t)(480-10t) to get the maximum first we find the derivative
f'(t)=-10(14+t)+480-10t
=-140-10t+480-10t to find the maximum we make the derivative=0 so
0=340-20t
20t=340
t=340/20
t=17
f(t)=(14+t)(480-10t) to get the maximum first we find the derivative
f'(t)=-10(14+t)+480-10t
=-140-10t+480-10t to find the maximum we make the derivative=0 so
0=340-20t
20t=340
t=340/20
t=17
-
let x = number of trees in excess of 14 , A = [ 14 + x ] [ 480 - 10 x ] , maximize A
{ 17 }
{ 17 }