Find Width and Length: Cuboid with Surface Area 300X cm² and Height 5X cm

Surface Area Applications with Parametric Variables

The surface area of a cuboid is 300X cm².

Its height is 5X cm.

What is its width and length?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:09 Let's begin by suggesting some options for the box's length and width.
00:15 Now, we'll use a formula to find the surface area of the box.
00:19 Let's plug in the right values from the data. Then, we'll solve for L, the length, and W, the width.
00:30 Next, simplify whatever you can.
00:51 Now, take out any common factors from the parentheses.
00:58 Alright, let's focus on finding the length L. We'll do this by isolating it.
01:14 Let's try setting the width, W, as 5.
01:21 Plug 5 into the equation wherever you see W.
01:45 Factorize 125 into smaller factors of 5 and 25.
01:51 Now, extract the common factor from inside the parentheses.
01:56 Simplify everything that's possible.
01:59 This will give us the length L when the width W is 5.
02:03 And that's how we find the solution to this problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

The surface area of a cuboid is 300X cm².

Its height is 5X cm.

What is its width and length?

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Apply the surface area formula to express an equation involving width and length.
  • Step 2: Substitute the known value of height (5X5X).
  • Step 3: Simplify and solve for the dimensions.

Now, let’s work through each step:

Step 1: The surface area formula for a cuboid is:

2(lw+lh+wh)=300X 2(lw + lh + wh) = 300X

Step 2: Substitute h=5Xh = 5X into the equation:

2(lw+5Xl+5Xw)=300X 2(lw + 5Xl + 5Xw) = 300X

Divide the entire equation by 2 to simplify:

lw+5Xl+5Xw=150X lw + 5Xl + 5Xw = 150X

Step 3: Simplify and solve for ll and ww:

To isolate one term, consider the equation:

lw+5X(l+w)=150X lw + 5X(l + w) = 150X

Let (w=5)(w = 5) and (l=25XX+1)(l = \frac{25X}{X + 1}), as per the calculations given in a choice example:

Therefore, we confirm this computation:

So, the width is 55 and the length is 25XX+1\frac{25X}{X + 1}.

As we check the answer choice, we agree that indeed the width and length meet the condition expressed in the given possible answer.

Width = 5

Height = 25XX+1\frac{25X}{X+1}

3

Final Answer

Width = 5

Height = 25xx+1 \frac{25x}{x+1}

Key Points to Remember

Essential concepts to master this topic
  • Formula: Use SA = 2(lw + lh + wh) for cuboid surface area
  • Technique: Substitute h = 5X into equation: 2(lw + 5Xl + 5Xw) = 300X
  • Check: Verify w = 5, l = 25X/(X+1) satisfies original equation ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting to factor out common terms with variables
    Don't leave lw + 5X(l + w) = 150X unsimplified = makes solving nearly impossible! Students get lost in complex algebra. Always factor out common variable terms and use strategic substitution to find specific solutions.

Practice Quiz

Test your knowledge with interactive questions

A cuboid is shown below:

222333555

What is the surface area of the cuboid?

FAQ

Everything you need to know about this question

Why can't I just solve for both width and length separately?

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Because you have one equation with two unknowns (width and length)! You need to make a strategic choice for one variable to find the other. The problem expects specific values like w = 5.

What does the X represent in this problem?

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X is a parameter - it's like a placeholder that makes the problem more general. Both the surface area (300X) and height (5X) contain this same variable, creating relationships between dimensions.

How do I know to choose w = 5?

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Look at the answer choices! They guide you toward reasonable values. Also, when you substitute w = 5 into your equation, it produces a clean expression for length: 25XX+1 \frac{25X}{X+1} .

Can there be other correct width and length combinations?

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Yes! This is actually a family of solutions. But the problem asks for specific values, so we choose the combination that matches the given answer format.

Why is my length expression so complicated?

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The length 25XX+1 \frac{25X}{X+1} looks complex but it's correct! Parametric problems often produce rational expressions like this. Don't worry - it simplifies for specific X values.

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