第12类RD Sharma解决方案–第18章，最大值和最小值–练习18.5 |套装1

问题1.确定两个正数，其和为15，且其平方和为最小值。

解决方案：

Let us assume the two positive numbers are x and y, 

And it is given that x + y = 15         …..(i)

So, let P = x² + y²                     …..(ii)

From eq (i) and (ii), we get

P = x² + (15 – x)²

On differentiating w.r.t. x, we get

dP/dx = 2x + 2(15 – x)(-1)

= 2x -30 +2x

= 4x -30

For maxima and minima.

Put dP/dx = 0

⇒ 4x – 30 = 0

⇒ x = 15/2

Since, d²P/dx² = 4 > 0

So, x = 15/2 is the point of local minima,

From eq(i), we get

 y = 15 – 15/2 = 15/2

So, the two positive numbers are 15/2, 15/2.

问题2。将64分为两部分，以使两部分的立方体的总和最小。

解决方案：

Let us assume 64 is divide into two parts that is x and y 

So, x + y = 64                    …..(i)

Let P = x³ + y³               ………(ii)

From eq(i) and (ii), we get

P = x³ + (64 – x)³

On differentiating w.r.t. x, we get

dP/dx = 3x² + 3(64 – x)² × (-1)

= 3x² – 3(4096 – 128x + x²)

= -3 (4096 – 128x)

For maxima and minima.

Put dP/dx = 0 

⇒ -3(4096 – 128x) = 0

⇒ x = 32

Now,

d²s/dx² = 384 > 0

So, x=32 is the point of local maxima.

Hence, the 64 is divide into two equal parts that is (32, 32)

问题3.我们应该如何选择两个大于或等于-2且总和为1/2的数字，以便使第一个和第二个立方的和最小。

解决方案：

Let us assume x and y be the two numbers, such that x, y ≥ -2 and

x + y = 1/2             ……(i)

So, let P = x + y³             …….(ii)

From eq(i) and (ii), we get

P = x + (1/2 – x)³

On differentiating w.r.t. x, we get

dP/dx = 1 + 3(1/2 – x)² × (-1)

= 1 – 3(1/4 – x + x²)

= 1/4 +3x -3x²

For maximum and minimum,

Put dP/dx = 0

⇒ 1/4 + 3x – 3x² = 0

⇒ 1 + 12x – 12x² = 0

⇒ 12x² – 12x – 1 = 0

⇒ $x=\frac{12±\sqrt{144+48}}{24}$

⇒ x = 1/2 ± (8√3/24)

⇒ x = 1/2 ± (1/√3)

⇒ x = {1/2 – (1/√3)}, {1/2 + (1/√3)}

Now,

d²P/dx² = 3 – 6x

So, at x =1/2 – (1/√3), d²P/dx² = 3(1 – 2(1/2 – 1/√3))

= 3(+2/√3) = 2√3 > 0

Hence, x = 1/2 – 1/√3 is point of local minima

From eq(i), we get

y = 1/2 – (1/2 – 1/√3) = 1/√3

So, the numbers are (1/2 – 1/√3) and 1/√3

问题4.将15分成两部分，使得一个的平方乘以另一个最小的立方。

解决方案：

Let us assume 15 is divide into two parts that is x and y 

So, x + y = 15

Also, P = x²y³

From eq(i) and (ii), we get

P = x²(15 – x)³

On differentiating w.r.t. x, we get

dP/dx = 2x(15 – x)³ – 3x²(15 – x)²

= (15 – x)²[30x – 2x² – 3x²]

= 5x(15 – x)²(6 – x)

For maxima and minima,

Put dP/dx = 0

⇒ 15(15 – x)²(6 – x) = 0

⇒ x = 0, 15, 6

Now,

So, d²P/dx² = 5(15 – x)²(6 – x) – 5x × 2(15 – x)(6 – x) –  5x(15 – x)²

At x = 0, d²P/dx² = 1125 > 0

So, x = 0 is point of local minima

At x = 15, d²P/dx² = 0

So, x = 15 is an inflection point.

At x = 6, d²P/dx² = -2430 < 0

So, x = 6 is the point of local maxima

So, the 15 is divide into two parts that are 6 and 9.

问题5.在所有封闭的圆柱形罐(右圆形)中，封闭的给定体积为100 cm ^3且具有最小的表面积吗?

解决方案：

Let us assume r be the radius of the cylinder and h be the height of the cylinder

So, the volume of the cylinder is 100 cm³

i.e., V = πr²h = 100

h = 100/πr²……(i)

Now we find the surface area of the cylinder is

A = 2πr² + 2πrh 

= 2πr² + 200/r

On differentiating w.r.t. r, we get

dA/dr = 4πr – 200/r², 

⇒ d²A/dr² = 4π + 400/r³

For maxima and minima,

dA/dr = 0 

⇒ 4πr = 200/r²

⇒ r³ = 200/4π = 50/π

r = (50/π)^1/3

So, when r = (50/π)^1/3, d²s/dr² > 0

Hence, from the second derivative test, the surface area is the minimum 

when the radius of the cylinder is (50/π)^1/3 cm

Now put the value of r in eq(i), we get

h = 100 / π(50/π)^1/3 = (2×50)/(50^2/3π^1-2/3) = 2(50/π)^1/3

问题6.梁的两端被支撑，并被均匀地加载。距一端的距离为x的弯矩M为

(1)M =(WL / 2)x –(w / 2)x ²

(ii)M = Wx / 3 –(W / 3)(x ³ / L ² )

找出每种情况下M的最大值。

解决方案：

(i) M = (WL/2)x – (w/2)x²

On differentiating w.r.t. x, we get

dM/dx = WL/2 – Wx

For maxima and minima,

Put dM/dx = 0 

WL/2 – Wx = 0   

x = L/2

Now, d²M/dx² = -W < 0

So, x = L/2 is point of local maxima.

Hence, M is maximum when x = L/2

(ii) M = Wx/3 – (W/3)(x³/L²)

On differentiating w.r.t. x, we get

dM/dx = W/3 – Wx²/L²

For maxima and minima,

Put dM/dx = 0 

W/3 – Wx²/L² = 0  

x = ± L/√3

Now, d²M/dx² = – 2xW/L²

So, at x = L/√3, ⇒ d²M/dx² =-2W/√3L < 0 (for max value)

at x = -L/√3, ⇒ d²M/dx² = 2W/√3L > 0 (for min value)

Hence, M is maximum when x = L/√3

问题7.长度为28 m的电线应切成两段。其中一件将被制成正方形，另一件将被制成圆形。两段的长度应该是多少，以使圆和正方形的总面积最小?

解决方案：

Let us assume l m be the piece of length cut from the given wire to make a square.

and  the other piece of wire that is used to create a circle is of length (28-l) m.

So, the side of square = l/4

Now, let us considered the radius of the circle is r. 

Then, 2πr = 28 – l ⇒ r = (1/2π)(28 – l)

Now we find the combined area of square and circle

A = l²/16 + π[(1/2π)(28 – l)]²

= l²/16 + 1/4π (28 – l)²

On differentiating w.r.t. l, we get

dA/dl = 2l/16 + (2/4π)(28 – l)(-1)

= l/8 – (1/2π)(28 – l)

d²A/dl² = l/8 + (1/2π) > 0

For maxima and minima,

Put dA/dl = 0 

⇒ l/8 – (1/2π)(28 – l) = 0

⇒ {πl – 4(28 – l)}8π = 0

⇒ (π + 4)l – 112 = 0

⇒ l = 112/(π + 4)

So, at l = 112/(π + 4), d²A/dl² > 0

Hence, using second derivative test, the area is the minimum when l = 112/(π + 4)

So, the length of the two pieces of wire are 112/(π + 4) and 28π/(π + 4) cm.

问题8.长度为20 m的电线应切成两段。一件将弯曲为正方形，另一件将弯曲为等边三角形。应该在哪里切割电线，以使正方形和三角形的面积之和最小?

解决方案：

According to the question

The length of the wire is 20 m

and the wire cut into two pieces x and y. So, x length wire is used to make a square and

y length wire is used to make triangle.

Now.

x + y = 20           …..(i)

x = 4l and y = 3a

So, A = sum of area of square and triangle

A = l² + √3/4a²         ……(ii)

We have, 4l + 3a  =20

4l = 20 – 3a

l = (20 – 3a) / 4

From eq(i), we have,

A = (20 – 3a)²/4 + √3/4a²

On differentiating w.r.t. a, we get

dA/da = 2{(20 – 3a)/4}(-3/4) + 2a × √3/4

For maxima and minima,

Put dA/da = 0

⇒ 2 {(20 – 3a)/4}(-3/4) + 2a × √3/4 = 0

⇒ -3(20 – 3a) + 4a√3 = 0

⇒ -60 + 9a + 4a√3 = 0

⇒ 9a + 4a√3 = 60

⇒ a(9 + 4√3) = 60

⇒ a = 60/(9 + 4√3)

Again differentiating w.r.t. a, we get

d²s/da² = (9 + 4√3)/8 > 0

So, the sum of the areas of the square and triangle is minimum when a = 60 / (9 + 4√3)

So l = (20 – 3a)/4

⇒ l = $\frac{20-3(60/(9+4√3))}{4}$

⇒ l = (180 + 80√3 – 180)/{4(9 + 4√3)}

⇒ l = 20√3/(9 + 4√3)

问题9.鉴于正方形和圆形的周长之和表明，当正方形的一侧等于圆形的直径时，它们的面积之和最小。

解决方案：

Let us assume the radius of the circle is r 

We have,

2πr + 4a = k (k is constant)

a = (k – 2πr)/4

Now we find the sum of the areas of the circle and the square:

A = πr² + a² = πr² + (k – 2πr)²/16

On differentiating w.r.t. r, we get

dA/dr = 2πr + 2(k – 2πr)(-2π)/16 

= 2πr- π(k – 2πr)/4

For maxima and minima,

Put, dA/dr = 0

2πr = π(k – 2πr)/4

8r = k – 2πr

r = k / (8 + 2π)= k / 2(4 + π)

Now, d²A/dr² = 2π + π²/2 > 0

So, at r = k / 2(4 + π), d²A/dr² > 0

Hence, the sum of the areas minimum when r = k / 2(4 + π)

So, a = $\frac{k-2π[\frac{k}{2}(4 + π)]}{4}$

= 2r 

Hence Proved

问题10。找到斜边长为5 cm的直角三角形的最大可能区域。

解决方案：

Let us assume PQR is a right-angled triangle, 

So, the hypotenuse h = PR = 5 cm.

Now, le us assume a and b be the remaining sides of the triangle. 

So, a² + b² = 25         ……(i)

Now we find the area of PQR = 1/2 QR × PQ

A = 1/2 ab             ……(ii)

From eq(i) and (ii), we get

⇒ A = 1/2 x √(25 – a²)

On differentiating w.r.t. a, we get

dA/da = $\frac{1}{2}[\frac{\sqrt{25-a^2}-2a^2}{2\sqrt{25-a^2}}]$

= $\frac{1}{2}[\frac{25-a^2-a^2}{\sqrt{25-a^2}}]$

= $\frac{1}{2}[\frac{25-2a^2}{\sqrt{25-a^2}}]$

For maxima and minima,

Put dA/da = 0

⇒ $\frac{1}{2}[\frac{25-2a^2}{\sqrt{25-a^2}}] =0$

⇒ a = 5/√2

Now,

d²A/d²a = $\frac{1}{2}[\frac{{\sqrt{25-a^2} (-4a)+\frac{(25-2a^2)2a}{2\sqrt{25-a^2}}}}{25-a^2}]$

At a = 5√2, d²s/d² = $\frac{1}{2}[\frac{\frac{-25}{√2}×5√2+0}{\frac{25}{2}}]$

= – 5/2 < 0

So, x = 5/√2 is a point local maxima,

Hence, the largest possible area of the triangle

= 1/2 × (5/√2) × (5/√2) = 25/4 square units

问题11.三角形的两个边长分别为“ a”和“ b”，它们之间的夹角为θ。 θ的哪个值将最大化三角形的面积?还要找到三角形的最大面积。

解决方案：

Let us assume ABC is a triangle such that AB = a, BC = b and ∠ABC = θ

and AD in perpendicular to BC.

BD = asinθ

So, the area of △ABC = 1/2 × BC × AD

⇒ A = 1/2 × b × a × sinθ

On differentiating w.r.t. θ, we get

dA/dθ = 1/2 × abcosθ

For maxima and minima,

Put dA/dθ = 0

⇒ 1/2 × abcosθ = 0

⇒ cosθ = 0

⇒ θ = π/2

Now, d²A/dθ² = -1/2 ab sinθ

At θ = π/2, d²A/dθ² = -1/2ab < 0

So, θ = π/2, is point of local maxima

Hence, the maximum area of the ABC triangle is 1/2 × absin(π/2) = 1/2 ab.

问题12.将一块18厘米长的方形锡盒制成一个没有顶部的盒子，方法是从每个角上切下一个正方形，并将折翼折叠成一个盒子。正方形的哪一面应被切除，以使盒子的体积最大?另外，找到此最大音量。

解决方案：

Let us assume that x cm be the side of the square to be cut off.

Now, the length and the breadth of the box will be (18 – 2x) cm each and 

the x cm be the height of the box.

So, the volume of the box is 

V (x) = x(18 – 2x)²

On differentiating w.r.t. x, we get

V'(x) = (18 – 2x)² – 4x(18-2x)

= (18 – 2x)[18 – 2x – 4x]

= (18 – 2x)(18 – 6x)

= 6 × 2(9 – x)(3 – x)

= 12(9 – x)(3 – x)

Again on differentiating w.r.t. x, we get

V”(x) = 12 [-(9 – x) – (3 – x)]

= -12 (9 – x + 3 – x)

= -12 (12 – 2x)

= -24 (6 – x)

For maxima and minima,

Put V'(x) = 0

12(9 – x)(3 – x) = 0

x = 9, 3

When x = 9 length and breadth of the box become zero.

So, x ≠ 9

When x = 3, V”(x) = -24 (6 – x) = -72 < 0

So, x = 3 is the point of maxima

Hence, the maximum volume is V_{x = 3} = 3(18 – 2 × 3)²

⇒ V = 3 × 12²

⇒ V = 3 × 144

⇒ V = 432 cm³ 

问题13.将一块45厘米乘24厘米的矩形锡纸制成一个没有顶部的盒子，方法是从每个角上切下正方形并折叠襟翼。正方形的哪一面应被切除，以使盒子的体积最大?

解决方案：

Let us assume that x cm be the side of the square to be cut off.

So, the height of the box = x, 

the length of the box = 45 – 2x, 

and the breath of the box = 24 – 2x.

So, the volume of the box is 

V(x) = x (45 – 2x)(24 – 2x)

= x (1080 – 90x – 48x + 4x²)

= 4x³ – 138x² + 1080x

On differentiating w.r.t. x, we get

V ‘(x)= 12x² – 276x + 1080

= 12(x² – 23x + 90)

= 12(x – 18) (x – 5)

Again on differentiating w.r.t. x, we get

V ”(x) = 24x – 276 = 12 (2x – 23)

For maxima and minima,

Put V'(x) = 0

4x³ – 138x² + 1080x = 0

x(x – 18) – 5(x – 18) = 0

(x – 5)(x – 18) = 0

So, x = 18 and x = 5

when x = 18 it is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. 

So, x ≠ 18

When x = 5, V ”(5) = 12 (10 – 23) = 12(-13) = -156 < 0

So, x = 5 is the point of maxima.

Hence, the volume of the box is maximum when x = 5.

问题14.建造一个底部为矩形且顶部为矩形的水箱，其深度应为2 m，容积为8 m ³ 。如果建造水箱的底价为每平方米₹70，侧面为每平方米₹45，那么最便宜的水箱的价格是多少?

解决方案：

Let us considered the length, breadth and height of the tank be l, b, and h

According to the question

The height of the tank is 2 and the volume is 8m³

So, the volume of the tank is  

V = l × b × h 

8 = l × b × 2

lb = 4  

⇒ b = 4/l    ….(i)

Now, we find the area of the base = lb = 4

and the area of the four walls (A) = 2h (l + b)

A = 4 (l + l/4)

On differentiating w.r.t. l, we get

⇒ dA/dl = 4 (l – 4/l²)

Again differentiating w.r.t. l, we get

 d²A/dl² = 32/l³

For maxima and minima,

Put dA/dl = 0

⇒ l – 4/l² = 0 

⇒ l² = 4

⇒ l = ±2

As we know that the length cannot be negative. So, l ≠ 2

When l = 2, d²A/dl² = 32/8 = 4 > 0

So, l = 2 is the point of minima.

Now put the value of l = 2 cm in the eq(i)

b = 4/l = 4/2 = 2

So, l = b = h = 2

Hence, the area is the minimum when l = 2.

So, the cost of building the base = 70 × (lb) = 70 × 4 = 280

Cost of building the walls = 2h (l + b) × 45 = 90 × 2 × (2 + 2) = 8 × 90 = 720

Hence, the total cost = 280 + 720 = 1000

问题15：矩形形式的窗口被一个半圆形的开口覆盖。窗户的总周长为10 m。找到窗口矩形部分的尺寸，以允许整个开口进入最大的光线。

解决方案：

Let us assume x and y be the length and breadth of the rectangle. 

So the radius of the semicircular opening = x/2

From the question it is given that the perimeter of the window is 10 m.

So, x + 2y + πx/2 = 10

⇒ x(1 + π/2) + 2y = 10

⇒ 2y = 10 – x(1 + π/2)

⇒ y = 5 – x(1/2 + π/4)

Now, the area of the window is

A = xy + (π/2)(x/2)²

= x [5 – x(1/2 + π/4)] + (π/8)x²

= 5x – x²(1/2 + π/4) + (π/8)x²

On differentiating w.r.t. x, we get

dA/dx = 5 – 2x(1/2 + π/4) + (π/4)x

= 5 – x(1 + π/2) + (π/4)x

Again differentiating w.r.t. x, we get

d²A/dx² = -(1 + π/2) + π/4 = -1 – π/4

For maxima and minima,

Put dA/dx = 0

⇒ 5 – x(1 + π/2) + (π/4)x = 0

⇒ 5 – x – (π/4)x = 0

⇒ x (1 + π/4) = 5

⇒ x = 5/(1 + π/4) = 20/(π + 4)

So, when x = 20/(π + 4), d²A/dx² < 0

So, the area is the maximum when length (x) = 20/(π + 4)

Now,

y = 5 – 20/(π + 4){(2 + π)/4} = 5 – 5(2 + π)/(π + 4) = 10/(π + 4) m 

Hence, the length of the rectangle is 20/(π + 4) m and the breadth is 10/(π + 4) m.