Question

An 8-inch diameter pizza stone costs $14.35.

Marco estimated that a 16-inch diameter pizza
stone (same thickness) should cost $28.70. Based
on the areas of the two pizza stones, is Marco's
estimate reasonable? Explain your answer. Please help! Thank you!

Answer 1

To determine if Marco's estimate is reasonable, we need to compare the areas of the two pizza stones.

The area of an 8-inch diameter pizza stone is:

A = πr^2
A = π(4)^2
A = 16π

The area of a 16-inch diameter pizza stone is:

A = πr^2
A = π(8)^2
A = 64π

To compare the areas, we can divide the area of the 16-inch pizza stone by the area of the 8-inch pizza stone:

64π / 16π = 4

This means that the area of the 16-inch pizza stone is four times greater than the area of the 8-inch pizza stone.

If we assume that the thickness of the two pizza stones is the same, then the cost of the 16-inch pizza stone should be proportional to its area:

Cost of 16-inch pizza stone / Cost of 8-inch pizza stone = Area of 16-inch pizza stone / Area of 8-inch pizza stone
Cost of 16-inch pizza stone / 14.35 = 64π / 16π
Cost of 16-inch pizza stone = (64π / 16π) x 14.35
Cost of 16-inch pizza stone = 4 x 14.35
Cost of 16-inch pizza stone = $57.40

Therefore, Marco's estimate of $28.70 for a 16-inch pizza stone is not reasonable, since the actual cost would be $57.40 based on the areas of the two pizza stones.

Answer 2

No, it is not reasonable.

Explanation:

Given an 8-inch pizza costs $14.35, you want to know if an estimated cost of $28.70 is reasonable for a 16-inch pizza.

Area

The scale factor for area is the square of the scale factor for linear dimensions such as diameter.

The ratio of diameters is ...

(16 in)/(8 in) = 2

Then the ratio of areas will be 2² = 4. The larger pizza will have 4 times the area of the smaller one.

If cost is based on area, then a better estimate of the cost of a 16-inch pizza is ...

4 × $14.35 = $57.40

Marco's estimate of $28.70 is not reasonable. A better estimate is $57.40.

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Additional comment

The area is given by the formula ...

A = (π/4)d²

When the diameter is doubled, the area becomes ...

A' = (π/4)(2d)² = 4×(π/4)d² = 4A

You can see that the scale factor 2 was squared when we expanded (2d)² to 2²·d² = 4d².

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