Find two positive real numbers whose product is a maximum.

puntgewelb5

puntgewelb5

Answered question

2021-11-11

Find two positive real numbers whose product is a maximum. The sum is 110.

Answer & Explanation

Witionsion

Witionsion

Beginner2021-11-12Added 19 answers

Let x and y be the two positive real numbers. Write an equation to show that the sum of the two positive real numbers is 110.
x+y=110
Solve for the value of y.
x+y=110
x+yx=110x
y=110x
Write an equation to show that the product of the two real numbers is the maximum. Let f(x) be the maximum. Substitute the values of x and y. Write as a quadratic function, f(x)=ax2+bx+c
f(x)=xy
f(x)=x(110x)
f(x)=110xx2
f(x)=x2+110x
a=1 which is a<0, the maximum at x=b2a. Detemine the value of x. Let a=-1 b=110
x=b2a
x=1102(1)
x=1102
x=55
Substitute the value of x in y=110x
y=110x
y=11055
y=55
The two positive real numbers are 55 and 55

nick1337

nick1337

Expert2023-05-25Added 777 answers

Step 1:
To find two positive real numbers whose product is a maximum and whose sum is 110, we can solve this problem using calculus. Let x and y be the two positive real numbers.
We are given the constraint that x+y=110. We need to find the values of x and y that maximize their product P=xy.
To solve this problem, we can use the method of Lagrange multipliers. We want to maximize the function P=xy subject to the constraint g(x,y)=x+y110=0.
The Lagrangian function is defined as L(x,y,λ)=xyλ(x+y110), where λ is the Lagrange multiplier.
To find the critical points, we take the partial derivatives of L with respect to x, y, and λ, and set them equal to zero:
Lx=yλ=0
Ly=xλ=0
Lλ=x+y110=0
From the first two equations, we can solve for x and y in terms of λ:
x=λ
y=λ
Step 2:
Substituting these values into the third equation, we have:
2λ110=0
Solving for λ, we get λ=55.
Using the values of λ obtained, we can find x and y:
x=λ=55
y=λ=55
Therefore, the two positive real numbers that maximize their product while summing up to 110 are x=55 and y=55.
To summarize, the solution is x=55 and y=55, with their product being P=xy=55·55=3025.
Vasquez

Vasquez

Expert2023-05-25Added 669 answers

Answer:
x=55 and y=55
Explanation:
The product of x and y can be expressed as P=xy. To find the maximum product, we can use the concept of the arithmetic-geometric mean inequality. According to this inequality, the arithmetic mean is always greater than or equal to the geometric mean.
The arithmetic mean of x and y is given by x+y2, and the geometric mean is given by xy. Therefore, we have the inequality:
x+y2xy
To maximize the product P, we want the left side of the inequality to be equal to the right side. Therefore, we can rewrite the inequality as an equation:
x+y2=xy
To solve this equation, let's square both sides:
(x+y2)2=(xy)2
Simplifying, we have:
x2+2xy+y24=xy
Multiplying both sides by 4, we get:
x2+2xy+y2=4xy
Rearranging the terms, we have:
x22xy+y2=0
Factoring, we obtain:
(xy)2=0
Taking the square root of both sides, we have:
xy=0
Solving for y, we find:
y=x
Since we are given that the sum of x and y is 110, we can write the equation:
x+x=110
Simplifying, we get:
2x=110
Dividing both sides by 2, we obtain:
x=1102
Hence, x=55.
Since y=x, we also have y=55.
Therefore, the two positive real numbers whose product is a maximum and whose sum is 110 are x=55 and y=55.
RizerMix

RizerMix

Expert2023-05-25Added 656 answers

Let x and y be the two positive real numbers. We want to find x and y such that their product, xy, is maximized while their sum, x+y, is equal to 110.
To solve this problem, we can use the method of Lagrange multipliers. We define the function:
L(x,y,λ)=xyλ(x+y110)
where λ is the Lagrange multiplier.
Next, we find the partial derivatives of L with respect to x, y, and λ and set them equal to zero:
Lx=yλ=0y=λ
Ly=xλ=0x=λ
Lλ=(x+y110)=0x+y=110
From the first two equations, we can see that x=y=λ. Substituting this into the third equation, we get:
2λ=110λ=55
Therefore, x=y=λ=55. The maximum product of two positive real numbers whose sum is 110 is x·y=55·55=3025.
Hence, the solution is x=y=55, and the maximum product is 3025.

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