Question

An artist’s canvas has sides measuring 3x – 3 and 4x + 1 inches.

What is the area of the canvas? Show all work.


The artist laid the canvas flat on the floor and poured some paint in the center. The paint flows at a rate of r(t) = 4t where t represents time in minutes and r represents how far the paint is spreading on the canvas. The area of the paint can be expressed as A[r(t)]= πr². What is the area of the circle created by the paint?


If the artist wants the circle to be at least 1300 in², will it be that large in 5 minutes? Support your answer with your work.

Answer 1

The length of the canvas is 3x - 3 inches and the width of the canvas is 4x + 1 inches. Therefore, the area of the canvas is:

A = (3x - 3)(4x + 1)
A = 12x² - 8x - 3

The paint is spreading at a rate of r(t) = 4t, which means that the radius of the circle is also increasing at a rate of 4t. Therefore, the area of the circle can be expressed as:

A[r(t)] = πr²
A[r(t)] = π(4t)²
A[r(t)] = 16πt²

We want to find out if the area of the circle will be at least 1300 in² in 5 minutes. So we need to substitute t = 5 into the equation for the area of the circle:

A[r(5)] = 16π(5)²
A[r(5)] = 400π

The area of the circle created by the paint after 5 minutes is 400π in², which is approximately 1256.64 in². Since this is less than 1300 in², the circle will not be that large in 5 minutes.

Answer 2

Explanation:

Finding the area of the canvas:

The canvas has sides measuring 3x - 3 and 4x + 1 inches. To find the area, we multiply the lengths of the sides. So, the area of the canvas is given by:

Area = (3x - 3) * (4x + 1)

= 12x^2 + x - 3

Therefore, the correct equation for the area of the canvas is A = 12x^2 + x - 3.

Finding the area of the circle created by the paint:

The rate at which the paint is spreading on the canvas is given by the function r(t) = 4t, where t represents time in minutes and r represents how far the paint has spread. The area of the circle created by the paint can be expressed as A[r(t)] = πr^2, where r(t) is the distance the paint has spread at time t.

Substituting the value of r(t) = 4t into the formula, we get:

A[r(t)] = π(4t)^2

= 16πt^2

Therefore, the area of the circle created by the paint is 16πt^2.

Determining if the circle will be at least 1300 in² in 5 minutes:

To check if the area of the circle will be at least 1300 in² in 5 minutes, we substitute t = 5 into the equation for the area:

A[r(5)] = 16π(5)^2

= 400π

Approximating the value of 400π, we get approximately, 1256.64.

Since 1256.64 is less than 1300, we can conclude that the area of the circle created by the paint after 5 minutes will not be at least 1300 in².