The problem is asking us to find the dimensions of a rectangle with the largest area, given that its perimeter is 188. Let's go step by step.

Given:

• Perimeter $P = 188$
• We need to maximize the area of the rectangle.

Step 1: Relationship between dimensions and perimeter

Let the dimensions of the rectangle be $l$ (length) and $w$ (width). The perimeter of a rectangle is given by: $P = 2l + 2w$ Substitute $P = 188$: $188 = 2l + 2w$ Dividing both sides by 2: $94 = l + w$ Thus, we have: $l = 94 - w$

Step 2: Area of the rectangle

The area $A$ of the rectangle is given by: $A = l \times w$ Substitute $l = 94 - w$ into the area formula: $A = (94 - w) \times w$ Expanding: $A = 94w - w^2$

Step 3: Maximizing the area

The area function $A = 94w - w^2$ is a quadratic equation. To find the maximum value of $A$, we find the vertex of the parabola, which occurs at: $w = \frac{-b}{2a}$ In the equation $A = -w^2 + 94w$, the coefficients are $a = -1$ and $b = 94$. So: $w = \frac{-94}{2 \times -1} = \frac{94}{2} = 47$

Step 4: Finding the dimensions

When $w = 47$, substitute this back into $l = 94 - w$ to find the length: $l = 94 - 47 = 47$

Thus, the rectangle with the largest area is a square with dimensions $47 \times 47$.

Step 5: Writing the answer as fractions

Since the dimensions are already whole numbers, the answer in fractional form (if necessary) would be: $\frac{47}{1} \times \frac{47}{1}$

Final Answer:

The dimensions of the rectangle with the largest area are $\frac{47}{1} \times \frac{47}{1}$ or simply $47 \times 47$.

Would you like more details on this? Here are some related questions to explore:

1. How do you determine the maximum or minimum of a quadratic function?
2. What is the significance of the fact that the largest-area rectangle is a square?
3. How can we use calculus to verify that $w = 47$ gives a maximum area?
4. How does the perimeter affect the shape of a rectangle with maximum area?
5. What are some practical applications of this type of optimization problem?

Tip: In optimization problems, if you're maximizing or minimizing a quantity, always check if the function you're working with is quadratic, as it will simplify finding the solution!