# Maximum area when height and width are dynamic?

At t = 0 the height is 8 and the length is 10. The length is decreasing at the rate of 1/4 of an inch per second while the height is increasing at the rate of half an inch per second. Is there a maximum possible area for the rectangle and if so when does it occur.

Just remembered how to do this. If we plot the area polynomial over time we get a parabola and we want to know when the parabola rate of change is zero, because that's when the parabola is at its max.

Therefore we take the derivative and set it equal to zero.

• Have you set up an equation for the area? – Toby Mak Feb 6 at 2:15
• What is the height as a function of time? What is the length as a function of time? Therefore, what is the area as a function of time? How do you then find the instant at which the area is maximum? – NickD Feb 6 at 2:16
• (10-1/4t)(8+1/2t) – Ole Feb 6 at 2:19
• @Ole Now expand, differentiate and set equal to 0. Also, please don't delete your last paragraph - it's quite useful to know you're just checking your answer. – Toby Mak Feb 6 at 2:20
• OK Cool - I'm just helping my kid out with some homework - It's been 20 years, so I'm a bit rusty on this. But I got it now - thank you. – Ole Feb 6 at 2:21

dA/dt=(10-t/4)(8+t/2). dA/dt=(10-t/4)(8+t/2).

• This is wrong: the area is the product of the length and the height, not the derivative of the area. To the OP, please uncheck this answer: it is not correct. – NickD Feb 6 at 15:16

We use the derivatives to find it.

• In contrast to your previous answer (which you should delete), this answer is correct. Please spend a little time to learn how to use MathJax to format mathematics on this site. – NickD Feb 6 at 15:18

The area, $$A$$ is $$A= LH.$$

Also, $$\frac{\mathrm{d}L}{\mathrm{d}t} = -\frac14$$ and $$\frac{\mathrm{d}H}{\mathrm{d}t} = \frac12$$.

$$\frac{\mathrm{d}A}{\mathrm{d}t} = L \frac{\mathrm{d}H}{\mathrm{d}t} + H \frac{\mathrm{d}L}{\mathrm{d}t} = \frac12 L - \frac14 H =\frac14(2L-H).$$

• Why $\frac 12 LH$? It's a rectangle, not a triangle. – Deepak Feb 6 at 2:35
• @user642005 +1 for including the general case. – Toby Mak Feb 6 at 11:16

So take the derivative of the area and set it equal to zero. The area is

$$A(t) = L\times H = (10 -\frac{1}{4}t) \times (8 + \frac{1}{2}t) = 80 + 3t -\frac{1}{8}t^2.$$

The derivative is:

$$A'(t) = 3 -\frac{1}{4}t.$$

Set the derivative equal to zero to find the (single) extremum:

$$3- \frac{1}{4}t = 0 \implies 3 = \frac{1}{4}t \implies t = 12.$$

The second derivative is negative, so the extremum is a maximum.

The maximum area is

$$A(12) = 80 + 3\cdot 12 - \frac{1}{8}(12)^2 = 80 + 36 - 18 = 98.$$