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Question Video: Finding the Dimensions of a Rectangular Body Using Differentiation Mathematics • Higher Education

The rectangular cross section of a block of wood is cut from a cylindrical log of diameter 67 cm. The resistance of this block is proportional to its width and the square of its length. What dimensions give the maximum resistance?

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Video Transcript

The rectangular cross section of a block of wood is cut from a cylindrical log of diameter 67 centimeters. The resistance of this block is proportional to its width and the square of its length. What dimensions give the maximum resistance?

In this example, we want to find the dimensions of a rectangular cross section of a block of wood, which is cut from a cylindrical log of diameter 67 centimeters. And we need these dimensions to give us the maximum resistance. To maximize the resistance, we need to cut the block so that the cross section diagonal is the same as the diameter of the log. If it was smaller, then the resistance would be lower.

So let’s begin by denoting the width of the cross section as 𝑀 and the length as 𝑙. The resistance, which we can denote as 𝑅, is a measure of the strength of the rectangular block, and it’s always positive. We’re told that the resistance is proportional to the width of the block 𝑀 and the square of its length 𝑙. This means that the resistance is proportional to the product of 𝑀 and 𝑙 squared. And so we have the relation 𝑅 is equal to π‘˜ times 𝑀 times 𝑙 squared, where π‘˜ is a positive real constant.

The width and length are also related to the diameter of the log by using the Pythagorean theorem on the right triangle. That is, 𝑀 squared plus 𝑙 squared is 67 squared, and that’s 4489. The optimization problem we want to solve is therefore maximize 𝑅 is equal to π‘˜ times 𝑀 times 𝑙 squared subject to the constraint 𝑀 squared plus 𝑙 squared is 4489.

Now, to find the maximum resistance, we’re going to differentiate 𝑅 with respect to 𝑀 to find the critical point or points and then use the second derivative test to classify them. And to do this, we must first rewrite 𝑅 as a function of 𝑀 only. And we can do this by rearranging our constraint as 𝑙 squared is equal to 4489 minus 𝑀 squared and substituting this into 𝑅 for 𝑙 squared. We then have 𝑅 as a function of 𝑀 only is equal to π‘˜ times 𝑀 times 4489 minus 𝑀 squared. And distributing our parentheses, that’s 4489 times π‘˜π‘€ minus π‘˜ times 𝑀 cubed.

Now, we can find our critical points by taking the first derivative of this function with respect to 𝑀 and setting it equal to zero. Differentiating 4489π‘˜π‘€ with respect to 𝑀 gives us 4489π‘˜. And differentiating negative π‘˜ times 𝑀 cubed with respect to 𝑀 using the power rule for derivatives, we multiply by the exponent and subtract one from the exponent. So we have negative three times π‘˜ times 𝑀 squared. And now setting this equal to zero, we solve for 𝑀.

Adding three π‘˜π‘€ squared to both sides, we can then divide through by π‘˜ and then divide through by three to give 4489 over three is 𝑀 squared. Then, taking the positive and negative square root on both sides, we have positive or negative 67 times the square root of three over three is equal to 𝑀. And since only nonnegative values of the width are valid, we can ignore the negative solution. And the only critical point is 67 times the square root of three over three is equal to 𝑀.

So now making some space and making a note of our critical point, to determine the nature of this critical point, we use the second derivative test, where the second derivative of 𝑅 with respect to 𝑀 is negative six π‘˜ times 𝑀, where the derivative of the constant 4489π‘˜ is equal to zero. And again, we’ve used the power rule to multiply by and subtract one from the exponent, giving negative six π‘˜π‘€.

The second derivative test tells us that for a twice differentiable function 𝑓 of π‘₯ and stationary or critical point π‘₯ sub zero of 𝑓 of π‘₯, if the second derivative with respect to π‘₯ at the point π‘₯ sub zero is greater than zero, then π‘₯ sub zero is a local minimum. If the second derivative of 𝑓 with respect to π‘₯ at the point π‘₯ sub zero is less than zero, then π‘₯ sub zero is a local maximum of 𝑓 of π‘₯. And if the second derivative at the point π‘₯ sub zero is equal to zero, then the test is inconclusive.

So now substituting our critical value 𝑀 is 67 root three over three into our second derivative, we have negative six π‘˜ multiplied by 67 root three over three. And since π‘˜ is a positive constant, our second derivative at the critical point is less than zero. And this means our critical point 𝑀 is 67 root three over three is a local maximum. That’s a local maximum of the resistance function 𝑅 of 𝑀.

We can find the length of the rectangular block corresponding to this value from the constraint. Recalling that we previously found that 𝑙 squared is 4489 minus 𝑀 squared, if we take the positive square root on both sides, since length is always positive, we have 𝑙 is equal to the square root of 4489 minus 67 root three over three squared. And this evaluates to the square root of 8978 over three, which in turn simplifies to 67 times the square root of six over three. And so making some space, we have the dimensions of the rectangular cross section that give the maximum resistance are 𝑀 is 67 root three over three centimeters and 𝑙 is 67 root six over three centimeters.

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