گاما رو نصب کن!

{{ number }}
اعلان ها
اعلان جدیدی وجود ندارد!
کاربر جدید

جستجو

پربازدیدها: #{{ tag.title }}

جستجوهای پرتکرار

میتونی لایو بذاری!

The diagram shows an open rectangular tank of height $h$ metres covered with a lid. The base of the tank has sides of length $x$ metres and $\frac{1}{2}x$ metres and the lid is a rectangle with sides of length $\frac{5}{4}x$ metres and $\frac{4}{5}x$ metres. When full the tank holds $4{\text{ }}{m^3}$ of water. The material from which the tank is made is of negligible thickness. The external surface area of the tank together with the area of the top of the lid is $A{\text{ }}{m^2}$.

a) Express $h$ in terms of $x$ and hence show that $A = \frac{3}{2}{x^2} + \frac{{24}}{x}$.

b) Given that $x$ can vary, find the value of $x$ for which $A$ is a minimum, showing clearly that $A$ is a minimum and not a maximum.

پاسخ تشریحی :
نمایش پاسخ

a) $h = \frac{8}{{{x^2}}}$

$A = \frac{1}{2}{x^2} + 2 \times \frac{1}{2}xh + 2xh + \frac{5}{4}x \times \frac{4}{5}x$

$A = \left( {3/2} \right){x^2} + 3xh$

$A = \frac{3}{2}{x^2} + 3x \times \frac{8}{{{x^2}}}$

$A = \frac{3}{2}{x^2} + \frac{{24}}{x}$

b) $\frac{{dA}}{{dx}} = 3x - \frac{{24}}{{{x^2}}} = 0$

$x = 2$

$\frac{{{d^2}A}}{{d{x^2}}} = 3 + \frac{{48}}{{{x^3}}}$

$ \gt 0$ when $x = 2$ hence minimum

تحلیل ویدئویی تست

تحلیل ویدئویی برای این تست ثبت نشده است!