Stupidly simple calculus problem with unexpected answer...

Question

Here is a simple calculus problem: The width of a rectangle is increasing at a rate of $\frac{1 \:cm} {min}$ and the height is decreasing at the rate $\frac{2 \:cm} {min}$. What is the rate of change of the area of a rectangle when it's width is $25cm$ and height is $15\:cm$ So, if $w$ is the width, $h$ is the height, $S$ is the area, the $\frac {dS}{dt}=dw\cdot h+dh \cdot w$= $1 \cdot 15+(-2) \cdot 25=15-50=-35$ The funny thing is that multiple choice for answers is as follows:

a)-2
b)-5
c)-1
d) 2

What am I doing wrong?... Thanks for looking.

I obtained the same result by setting $h= 15-2t$ and $w=25+t$. — Vladimir Vargas, Nov 16 '14 at 17:26
Or probably the problem says: the width decreases at a rate $2$cm/min and the height increases at a rate $1$cm/min. This will give the answer b). — Vladimir Vargas, Nov 16 '14 at 17:28
Nope, the problem is exactly as I printed here... I've also noticed that if I switch h=25, w=15, I'd get -5, but I can't change condition of the problem just to get one of the "correct" answers... — user175755, Nov 16 '14 at 17:31
They probably messed up the problem when they were writing it. — Vladimir Vargas, Nov 16 '14 at 18:05

yess · Accepted Answer · 2014-11-16T18:25:13.420

1

Trust your mathematics.

You have a function of the area in terms of time given by

$$S(t)=h(t)w(t)$$

then you differentiate and obtain

$$S'(t)=h'(t)w(t)+h(t)w'(t)$$

using the suggestion of Vladimir Vargas for $h=15−2t$ and $w=25+t$. You obtain $$S'(0)=-35$$ .

edited Nov 16 '14 at 18:25

answered Nov 16 '14 at 17:39

yess

1,002

1

Many thanks to Vladimir Vargas and Yess! – user175755 Nov 16 '14 at 17:43

Stupidly simple calculus problem with unexpected answer...

1 Answers1