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A rectangular storage container without a lid is to have a volume of 10 m3. The length of its base is twice the width. Material for the base costs $15 per square meter. Material for the sides costs $9 per square meter. Let w denote the width of the base. Find a function in the variable w giving the cost C (in dollars) of constructing the box. C(w) = Find the derivative of C. C'(w) = Find the cost (in dollars) of materials for the least expensive such container. (Round your answer to the nearest cent.)

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MrRoyal

The cost function is C(w) = 300/w + 18w² and the cost of materials for the cheapest such container is $222.0

The function for constructing the cost of the box

From the question, we have the following parameters that can be used in our computation:

The length is twice the width;

Volume = 10m³

This means that:

V = 10

l = 2w

The volume of the container is calculated as:

V = lwh

So, we have

v = 2w²h

By substitution, we have

h = v/2w²

So, we have

h = 10/2w²

Divide

h = 5/w²

The surface area of an open rectangular storage is:

A = 2h(l + w) + lw

So, we have

A = 2h(2w + w) + 2w²

A = 2h(3w) + 2w²

Substitute h = 5/w²

A = 2 * 5/w² * (3w) + 2w²

Evaluate

A = 20/w + 2w²

The base costs $15, and the sides cost $9

So, the cost function is:

C(w) = 15 * 20/w + 2w² * 9

Rewrite as:

C(w) = 300/w + 18w²

The least cost of the materials

We have

C(w) = 300/w + 18w²

Differentiate

C'(w) = -300/w² + 36w

Set to 0

-300/w² + 36w = 0

Rewrite as:

-36w = -300/w²

Cancel out the negatives

36w = 300/w²

Multiply through by ²

36w³ = 300

Divide through by 36

w³ = 8.33

Take cube roots of both sides

w = 2.03

Recall that:

C(w) = 300/w + 18w²

So, we have:

C(2.03) = 300/2.03 + 18(2.03)²

Evaluate

C(2.03) = 222.0

Hence, the cost of the materials is $222.0

Read more about minimizing functions at:

brainly.com/question/12313753

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