All 12 Optimization Problems Solved Beautifully
Find two positive numbers whose sum is 12 and their product is maximum.
Let the two numbers be x and y.
Given that their sum is 12:
x + y = 12
We can express y in terms of x:
y = 12 - x
The product P of the two numbers is:
P = x × y = x(12 - x) = 12x - x²
This is a quadratic function that opens downward (since coefficient of x² is negative).
The maximum occurs at the vertex. For a function ax² + bx + c, the vertex is at x = -b/(2a).
Here, a = -1, b = 12
x = -b/(2a) = -12/(2×-1) = 6
When x = 6:
y = 12 - 6 = 6
So both numbers are 6.
P = 6 × 6 = 36
The two numbers are 6 and 6, giving the maximum product of 36.
Graph showing how the product changes with different numbers:
The tallest bar shows the maximum product occurs when both numbers are equal.
Find two positive numbers whose product is 20 and their sum is minimum.
Let the two numbers be x and y.
Given that their product is 20:
x × y = 20
We can express y in terms of x:
y = 20/x
The sum S of the two numbers is:
S = x + y = x + 20/x
Take derivative of S with respect to x:
dS/dx = 1 - 20/x²
Set derivative to zero for critical points:
1 - 20/x² = 0 ⇒ x² = 20 ⇒ x = √20 ≈ 4.47
When x = √20:
y = 20/√20 = √20
So both numbers are √20.
S = √20 + √20 = 2√20 ≈ 8.94
The two numbers are √20 and √20 (≈4.47), giving the minimum sum of 2√20 ≈ 8.94.
Graph showing how the sum changes with different numbers:
The shortest bar shows the minimum sum occurs when both numbers are equal.
Find the smallest possible value of x² + y² given that x + y = 10.
x + y = 10 ⇒ y = 10 - x
S = x² + y² = x² + (10 - x)² = x² + 100 - 20x + x² = 2x² - 20x + 100
This is a quadratic function opening upward (a=2 > 0).
Minimum occurs at vertex: x = -b/(2a) = 20/4 = 5
y = 10 - 5 = 5
S = 5² + 5² = 25 + 25 = 50
The minimum value is 50 when x = y = 5.
Graph of the sum of squares function:
The curve shows the minimum sum of squares occurs when x=5.
A garden is to be laid out in a rectangular area and protected by wire fence. What is the largest possible area of the fenced garden with 40 metres of wire?
Let length = x meters, width = y meters.
2(x + y) = 40 ⇒ x + y = 20 ⇒ y = 20 - x
A = x × y = x(20 - x) = 20x - x²
Quadratic opening downward (a=-1), vertex at:
x = -b/(2a) = -20/(2×-1) = 10
y = 20 - 10 = 10
Both length and width are 10m (a square).
A = 10 × 10 = 100 m²
The maximum area is 100 m² with a square garden (10m × 10m).
Different rectangle shapes with perimeter 40m:
8m × 12m
Area: 96m²
9m × 11m
Area: 99m²
10m × 10m
Area: 100m² (MAX)
The square gives the maximum area for a given perimeter.
A rectangular page is to contain 24 cm² of print. The margins at the top and bottom of the page are 1.5 cm and the margins at other sides of the page is 1 cm. What should be the dimensions of the page so that the area of the paper used is minimum.
Let print area dimensions be width = x cm, height = y cm.
x × y = 24 ⇒ y = 24/x
Total page width: x + 2 (1cm margins on both sides)
Total page height: y + 3 (1.5cm margins top and bottom)
A = (x + 2)(y + 3) = (x + 2)(24/x + 3) = 24 + 3x + 48/x + 6 = 30 + 3x + 48/x
Take derivative:
dA/dx = 3 - 48/x²
Set to zero:
3 - 48/x² = 0 ⇒ x² = 16 ⇒ x = 4
y = 24/4 = 6
Total width: 4 + 2 = 6 cm
Total height: 6 + 3 = 9 cm
The optimal page dimensions are 6cm × 9cm.
Page layout with margins:
Margins: 1.5cm top/bottom, 1cm sides
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq.mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Let side parallel to river = x meters
Side perpendicular to river = y meters
x × y = 180,000 ⇒ y = 180,000/x
Only 3 sides need fencing (no river side):
L = x + 2y = x + 2(180,000/x) = x + 360,000/x
Take derivative:
dL/dx = 1 - 360,000/x²
Set to zero:
1 - 360,000/x² = 0 ⇒ x² = 360,000 ⇒ x = 600
y = 180,000/600 = 300
L = 600 + 2×300 = 1200 meters
The minimum fencing needed is 1200 meters (600m parallel to river and 300m on each side).
Pasture layout:
Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10 cm.
A rectangle inscribed in a circle means all four vertices lie on the circle.
This implies the diagonal of the rectangle is the diameter of the circle.
Let the rectangle have length l and width w.
The diagonal d = √(l² + w²) = 2×10 = 20 cm (diameter)
l² + w² = 400
We can express w in terms of l:
w = √(400 - l²)
A = l × w = l × √(400 - l²)
Using calculus, we find maximum occurs when l = w (square).
From l² + l² = 400 ⇒ 2l² = 400 ⇒ l² = 200 ⇒ l = √200 ≈ 14.14 cm
w = √(400 - 200) = √200 ≈ 14.14 cm
The rectangle with maximum area is a square with sides ≈14.14 cm (√200 cm).
Circle with inscribed rectangle:
The largest inscribed rectangle is a square with diagonal equal to the diameter.
Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Let perimeter P = 2(l + w) be constant.
Let l + w = k (constant), so w = k - l
A = l × w = l(k - l) = kl - l²
This is a quadratic function opening downward.
Maximum occurs at vertex: l = -b/(2a) = -k/(2×-1) = k/2
w = k - k/2 = k/2
Thus l = w, meaning the rectangle is a square.
A = (k/2) × (k/2) = k²/4
For a fixed perimeter, the square (where length = width) gives the maximum area among all rectangles.
Rectangles with perimeter P=40:
5×15
Area: 75
10×10
Area: 100 (MAX)
15×5
Area: 75
The square has the largest area for the same perimeter.
Find the dimensions of the largest rectangle that can be inscribed in a semicircle of radius r cm.
The rectangle sits on the diameter with its top corners touching the semicircle.
Let the rectangle have width 2x (from -x to x on diameter) and height y.
The top corner (x,y) lies on the semicircle: x² + y² = r² ⇒ y = √(r² - x²)
A = 2x × y = 2x√(r² - x²)
Using calculus, maximum occurs when x = r/√2
Then y = √(r² - r²/2) = r/√2
Width: 2x = 2r/√2 = r√2
Height: y = r/√2
The largest rectangle has dimensions width = r√2 and height = r/√2.
Semicircle with inscribed rectangle:
The optimal rectangle has width √2 times its height.
A manufacturer wants to design an open box having a square base and a surface area of 108 sq.cm. Determine the dimensions of the box for the maximum volume.
Let base side = x cm, height = h cm.
Open box has 1 base + 4 sides:
x² + 4xh = 108 ⇒ h = (108 - x²)/(4x)
V = x² × h = x² × (108 - x²)/(4x) = (108x - x³)/4
Take derivative:
dV/dx = (108 - 3x²)/4
Set to zero:
108 - 3x² = 0 ⇒ x² = 36 ⇒ x = 6
h = (108 - 36)/24 = 72/24 = 3 cm
The box should have dimensions 6cm × 6cm × 3cm for maximum volume.
Open box with square base:
Base: 6×6, Height: 3cm, Surface Area: 36 + 4×18 = 108cm²
The volume of a cylinder is given by the formula V = πr²h. Find the greatest and least values of V if r + h = 6.
r + h = 6 ⇒ h = 6 - r
V = πr²h = πr²(6 - r) = 6πr² - πr³
Take derivative:
dV/dr = 12πr - 3πr² = 3πr(4 - r)
Set to zero:
3πr(4 - r) = 0 ⇒ r = 0 or r = 4
At r = 0: V = 0 (minimum)
At r = 4: h = 2, V = π×16×2 = 32π (maximum)
The maximum volume is 32π when r=4 and h=2.
The minimum volume is 0 when r=0 or h=0.
Volume as function of radius:
The volume reaches its maximum at r=4 then decreases.
A hollow cone with base radius a cm and height b cm is placed on a table. Show that the volume of the largest cylinder that can be hidden underneath is 4/9 times volume of the cone.
We need to find the largest cylinder that fits entirely within the cone.
Let cylinder radius = r, height = h.
By similar triangles: (b-h)/b = r/a ⇒ r = a(1 - h/b)
V = πr²h = πa²(1 - h/b)²h = πa²(h - 2h²/b + h³/b²)
Take derivative:
dV/dh = πa²(1 - 4h/b + 3h²/b²)
Set to zero:
3h²/b² - 4h/b + 1 = 0 ⇒ h = b/3 or h = b
The valid solution is h = b/3 (h=b gives zero volume)
r = a(1 - (b/3)/b) = 2a/3
Cylinder volume: V = π(2a/3)²(b/3) = (4πa²b)/27
Cone volume: V_cone = (πa²b)/3
Ratio: V/V_cone = (4/27)/(1/3) = 4/9
The largest cylinder has volume (4/9) times the volume of the cone, proved.
Cone with inscribed cylinder:
The optimal cylinder has height 1/3 and radius 2/3 of the cone's dimensions.