bookishfairy73
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A construction engineer needs to make 25 square concrete slabs with sides of length x feet and a height x - 3 feet. If the engineer can use at most 350,000ft^3 of concrete, what should the dimensions of each slab be, to the nearest foot?

a. Write an expression for the volume of all the slabs

b. What should x equal, to the nearest foot?

Verified Answer :

We know that the given side lengths are x and the height is x. 

Therefore, the area of the base =  \(A = x * x\) 

\(Formula =\ \textgreater \ volume = length*width*height\)

\(To \ find \ the \ volume \ of \ 1 \ slab, \ multiply \ by \ the \ height\) \(x -3\) 

\(= x*x*(x-3)\) 

\(Volume \ of \ 25 \ slabs =\) \(25*x^2*(x-3)=350000\) 

\(350,000/25 = 14,000\)

\(x^3 - 3x^2 - 14000\)

\(= 25.14\)

\(Rounded => x=25\)


maqibkhan234

Answer:

x ≈ 118 ft

Step-by-step explanation:

To get an expression for the volume of all slabs, we can use the formula of volume of cuboid 'V', which is equal to the product of its length 'x', breadth 'x' and height 'x-3' i.e.

\(V = x.x.(x-3)\)

For 25 slabs,

\(V = 25x.x.(x-3)\)

Since, V cannot be more than 350,000 \(ft^{3}\), therefore,

\(350000 = 25x^{2} (x-3)\)

which can be written as:

either:

\(350000 = 25x^{2}\)

\(x = \sqrt{ \frac{350000}{25}\)

x ≈ 118 ft

or:

\(x - 3 = 350000\)

\(x = 350003\)

which is not possible.

Hence x should be 118 ft

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