Question

Для строительства магазина заготовлен материал на наружные стены длиной 24 м и высотой 3 м. Какими должны быть размеры магазина (в виде прямоугольного параллелепипеда), чтобы он имел наибольший объём? Ответ: .​

Answer 1

Ответ:

6*6*3 м

Объяснение:

24*3=48 м^2
найбільший обєм, коли в основі квадрат
4*a=24
a=24:4=6 м
тоді розміри
6*6*3 м

Answer 2

Calculating the Maximum Volume of the Store

To calculate the maximum volume of the store, we can use the given material for the outer walls, which has a length of 24m and a height of 3m. We need to find the dimensions of the store in the form of a rectangular parallelepiped that would result in the largest possible volume.

Let's calculate the dimensions of the store that would result in the largest volume.

Solution:

To maximize the volume, we can use the given material for the outer walls to construct a rectangular parallelepiped with the largest possible volume. The volume of a rectangular parallelepiped is given by the formula: V = lwh, where l is the length, w is the width, and h is the height.

Given: - Length of material for outer walls = 24m - Height of material for outer walls = 3m

We want to find the dimensions of the store (length, width, and height) that would result in the largest volume.

Calculating the Width:

To maximize the volume, we can consider the width as x meters. This means the length of the material will be used for the perimeter of the base of the store, and the height of the material will be used for the height of the store.

Calculating the Volume:

Using the formula for the volume of a rectangular parallelepiped, V = lwh, we can express the volume in terms of x:

V = 24x * 3 * x = 72x^2

To find the value of x that maximizes the volume, we can take the derivative of the volume function with respect to x and set it equal to 0.

Derivative and Maximum Volume:

Taking the derivative of the volume function with respect to x: dV/dx = 144x

Setting the derivative equal to 0 to find the critical points: 144x = 0

Solving for x: x = 0

However, since x represents the width of the store, it cannot be 0. Therefore, the critical point does not apply in this context.

Conclusion:

The store should be constructed as a rectangular parallelepiped with the dimensions that maximize the volume, which would be 24m in length, x meters in width, and 3m in height. This would result in the store having the largest possible volume.

This calculation assumes that the material for the outer walls is used efficiently to construct the store with the given dimensions.