07/03/2018
Examples of Maxima/Minima Problems with Solutions.
1. Given that x+y=7, find the minimum value of the expression x^2+7y^2 provided that x, y > 0.
Solution:
Since y=7-x, we have to find the minimum of x^2+7(7-x)^2.
Let f(x) = x^2 + 7(7-x)^2
f(x) = x^2 + 7(7-x)^2
Taking f'(x) we have,
f'(x) = 2x + 14(x-7)
f'(x) = 2x + 14x - 98
f'(x) = 16x - 98
Equating f'(x) = 0,
16x - 98 = 0
16x = 98
x = 49/8
Recall y=7-x,
y = 7 - 49/8
y = 7/8
Therefore, the minimum of x^2 + 7y^2 is,
(49/8)^2 + 7(7/8)^2 = 343/8
2. Find two positive real numbers whose product is 121 and whose sum is a minimum.
Solution:
Let x -> 1st number
y --> 2nd number
xy = 121
y = 121/x
Minimizing x+y we have,
f(x) = x + y
f(x) = x + 121/x
Taking f'(x) we have,
f'(x) = 1 - 121/x^2
Letting f'(x) = 0,
1 - 121/x^2 = 0
121/x^2 = 1
x^2 = 121
x = 11
Recall xy=121,
11y = 121
y = 11
Therefore, the numbers are 11 and 11.
3. Find two positive real numbers x and y such that x + y = 20 and x^3 + y^2 has a minimum value.
Solution:
Since,
x + y = 20
y = 20 - x
Substituting y to x^3 + y^2,
f(x) = x^3 + (20-x)^2
Taking the derivative of f,
f'(x) = 3x^2 + 2(x-20)
f'(x) = 3x^2 + 2x - 40
Letting f'(x) = 0,
3x^2 + 2x - 40 = 0
(3x - 10)(x + 4) = 0
x = 10/3, x = - 4
Taking f''(x),
f''(x) = 6x + 2
Substiuting x = 10/3 yields a positive number while x = -4 yields a negative number. Therefore, x = 10/3 yields a minimum.
Recall x + y = 20,
10/3 + y = 20
y = 20 - 10/3
y = 50/3
Therefore, the two numbers are 10/3 and 50/3 respectively.
4. It is found that if a hectare of land is planted with 42 mango trees, the annual yield of each tree will be 480 mangoes. For every additional tree planted on this land, the yield per tree will decrease by 8 mangoes. How many trees should be planted to this hectare of land to maximize the annual yield?
Solution:
Let Y(x) be the quadratic function that defines the relation.
Y(x) = (42 + t)(480 - 8t)
Y(x) = -8t^2 + 144t + 20160
Taking the derivative of Y,
Y'(x) = -16t + 144
Letting Y'(x) = 0,
-16t = -144
t = 9
Therefore, there should be 42 + 9 = 51 trees that should be planted in order to maximize the yield.
5. The hypotenuse of a right triangle has length 14 m. Find its perimeter if its area is maximum.
Solution:
By the Pythagorean Theorem,
a^2 + b^2 = c^2
a^2 + b^2 = 14^2
a^2 + b^2 = 196
b^2 = 196 - a^2
b = √(196 - a^2)
A = (1/2)(ab)
A = (1/2)(a)√(196 - a^2)
A = (1/2)√(196a^2 - a^4)
Let g(a) = 196a^2 - a^4
g(a) = 196a^2 - a^4
g'(a) = 392a - 4a^3
Letting g'(a) = 0,
392a - 4a^3 = 0
a(392 - 4a^2) = 0
4a^2 = 392
a^2 = 98
a = 7√(2)
Recall,
a^2 + b^2 = 196
98 + b^2 = 196
b^2 = 98
b = 7√(2)
Therefore, the perimeter of the triangle is
P = 14√(2) + 14 m
