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Question
Math
Posted 5 months ago

Squares of width xx are removed from a 20 - cm\mathrm{cm} by 27 - cm\mathrm{cm} piece of cardboard, and the resulting edges are folded up to form a box with no top. Determine all values of xx so that the volume of the resulting box is at most 450 cm3450 \mathrm{~cm}^{3}.

The values of xx are
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Answer from Sia
Posted 5 months ago
Solution by Steps
step 1
We start by setting up the equation for the volume of the box. The volume VV of the box is given by the product of its length, width, and height. The length and width of the box are reduced by 2x2x each due to the squares cut out from each corner, and the height of the box is xx. Therefore, the volume VV is given by: V=(202x)(272x)x V = (20 - 2x)(27 - 2x)x
step 2
We need to find the values of xx such that the volume VV is at most 450cm3450 \, \text{cm}^3. This gives us the inequality: (202x)(272x)x450 (20 - 2x)(27 - 2x)x \leq 450
step 3
We solve the inequality: (202x)(272x)x450 (20 - 2x)(27 - 2x)x \leq 450 Expanding the left-hand side: (202x)(272x)x=(54040x54x+4x2)x=4x394x2+540x (20 - 2x)(27 - 2x)x = (540 - 40x - 54x + 4x^2)x = 4x^3 - 94x^2 + 540x So the inequality becomes: 4x394x2+540x450 4x^3 - 94x^2 + 540x \leq 450
step 4
We simplify the inequality: 4x394x2+540x4500 4x^3 - 94x^2 + 540x - 450 \leq 0
step 5
We solve the cubic inequality. First, we find the roots of the equation: 4x394x2+540x450=0 4x^3 - 94x^2 + 540x - 450 = 0 Using numerical methods or a graphing calculator, we find the roots to be approximately: x1,x7.5,x15 x \approx 1, \quad x \approx 7.5, \quad x \approx 15
step 6
We determine the intervals where the inequality holds. Testing intervals around the roots, we find that: 1x7.5 1 \leq x \leq 7.5
Answer
The values of xx are 1x7.51 \leq x \leq 7.5
Key Concept
Solving a cubic inequality
Explanation
To determine the values of xx that satisfy the volume constraint, we set up the inequality, expand and simplify it, find the roots, and test the intervals around the roots.

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